To solve this problem it is necessary to apply the concepts related to momentum theorem.
The equation for impulse is given as
[tex]I = Ft[/tex]
Where
I = Force
t = Time
At the same time we have the equation for momentum is given as
[tex]p = mv[/tex]
The impulse momentum theorem states that the change in momentum of an object is equal to the impulse applied to it. Therefore
I = p
Ft = mv
Solving to find the force
[tex]F = \frac{mv}{t}[/tex]
[tex]F = \frac{(70)(6)}{0.25}[/tex]
[tex]F = 1680N[/tex]
Therefore the magnitude of the average horizontal force by diver on the raft is 1680N
The magnitude of the average horizontal force exerted by the diver on the raft is 1680 N.
Explanation:To find the magnitude of the average horizontal force exerted by the diver on the raft, we need to start by calculating the change in momentum of the diver. The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. The change in momentum is equal to the impulse, which is given by the equation J = Δp = mΔv.
Since the swimmer dives horizontally, the change in velocity is equal to the initial velocity of the swimmer. Therefore, Δv = 6.0 m/s. Substituting the values, we get J = (70 kg)(6.0 m/s) = 420 kg·m/s.
The impulse is equal to the average force multiplied by the time interval, so we can rearrange the equation to solve for the average force. F = J / Δt = 420 kg·m/s / 0.25 s = 1680 N.
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A wooden artifact is found in an ancient tomb. Its 14C activity is measured to be 66.3% of that in a fresh sample of wood from the same region. Assuming the same amount of 14C was initially presented in the wood from which the artifact was made, determine the age of the artifact. The half-life of 14C is 5730 y. Answer in units of y.
Answer:
3396.53 years
Explanation:
Using decay formula
In([tex]\frac{N}{No}[/tex]) = -Kt where t is the age of the artifact in years and k is the decay constant
T1/2 = [tex]\frac{In2}{K}[/tex]
5730 = [tex]\frac{In2}{K}[/tex]
K = In 2 / 5730= 0.000121yr^-1
N / No = 0.663
In (0.663) / -0.000121 = t
t = 3396.53 years
Since we cannot physically collect data from stars and most other objects in the universe, almost all of the information we obtain from the universe comes from analyzing the light, or spectra, from those objects. The study of light is known as spectroscopy. As we have seen in this simulation, every blackbody emits light with an easily identified pattern known as the blackbody curve. This is the particular way the total light emitted by a blackbody varies with its frequency. The exact form of the curve depends only on the body's temperature. Since we can treat stars as blackbodies, this is incredibly useful in astronomy that shows us that the color of a star is also indicative of its temperature. Use the simulation to determine the surface temperature of the following star: Betelgeuse is a red supergiant star in the constellation Orion. Knowing that Betelgeuse has peak intensity in the red and infrared wavelengths, adjust the intensity scale and temperature until you can determine the approximate surface temperature of the star. a. 3500K b. 4800K c. 7700K d. 11,000 K
Answer:
3500 K
Explanation:
b = Wien's displacement constant = [tex]2.89\times 10^{-3}\ mK[/tex]
Wavelength range = 700 nm to 10⁶ m. Let us take 825 nm
[tex]\lambda_m=825\ nm[/tex]
From Wien's displacement law we have
[tex]\lambda_m=\dfrac{b}{T}\\\Rightarrow T=\dfrac{b}{\lambda_m}\\\Rightarrow T=\dfrac{2.89\times 10^{-3}}{825\times 10^{-9}}\\\Rightarrow T=3500\ K[/tex]
The surface temperature of Betelguese is 3500 K
A horse running at 3 m/s speeds up with a constant acceleration of 5 m/s2. How fast is the
horse going when it travels 15.3 m from where it started to accelerate.
Answer:
The horse is going at 12.72 m/s speed.
Explanation:
The initial speed of the horse (u) = 3 m/s
The acceleration of the horse (a)= 5 m/[tex]s^{2}[/tex]
The displacement( it is assumed it is moving in a straight line)(s)= 15.3 m
Applying the second equation of motion to find out the time,
[tex]s=ut+\frac{1}{2}at^{2}[/tex]
[tex]15.3=3t+2.5t^{2}[/tex]
[tex]2.5t^{2}+3t-15.3=0[/tex]
Solving this quadratic equation, we get time(t)=1.945 s, the other negative time is neglected.
Now applying first equation of motion, to find out the final velocity,
[tex]v=u+at[/tex]
[tex]v=3+1.945*5[/tex]
[tex]v=3+9.72[/tex]
v=12.72 m/s
The horse travels at a speed of 12.72 m/s after covering the given distance.
The SI unit of power is the watt. Which of the following units are equivalent to the watt?
A) V∙AB) J/CC) C/sD) V/sE) A/s
Answer:
The right option is (A) V.A
Explanation:
Power: This is the rate at which work is done. Or it is the produce of force and velocity. The S.I unit of power is Watt (W). Other units include Horse power(hp), foot-pound per minutes, etc.
Generally, power can be represented as,
Power = Energy/time
P = W/t......................... Equation 1
Where p = power, w = Work or energy, t = time in seconds.
Electrical energy: This is the product of potential difference and the quantity of charge.
∴ W = VQ............................... Equation 2
Where V = potential difference, Q = quantity of charge and W = Energy or Work done.
Also Q = It........................ Equation 3.
where I = current in ampere, t = time in seconds
Substituting equation 3 into equation 3
W = VIt............................ Equation 4.
Also substituting Equation 4 into Equation 1
P = VIt/t = VI = voltage(V)×Current(A)
Therefore the equivalent unit of power is
P = V.A.
The right option is (A) V.A
The SI unit of power is the watt. The options A) V∙A, B) J/C, and C) C/s are equivalent to a watt.
Explanation:The SI unit of power is indeed the watt, represented by the symbol 'W'. The watt is a derived unit of power in the International System of Units (SI) and is defined as one joule per second. Hence, three of the given options, A) V∙A, B) J/C, and C) C/s are equivalent to a watt.
A) A Volt (V) times an Ampere (A) also equals a watt (V∙A=W). This is derived from the formula P=V∙I where P is power, V is voltage, and I is current. B) A Joule (J) per Coulomb (C) is also a watt (J/C = W). This comes from the relationship P=W/t = J/s = V∙A. C) A Coulomb (C) per second is also a watt (C/s = W), because one ampere equals to 1 C/s. The unit Coulomb per second refers to the electrical current where 1 A is equivalent to 1 C/s.Learn more about Watt here:https://brainly.com/question/27355276
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When radio waves try to pass through a city, they encounter thin vertical slits: the separations between the buildings. This causes the radio waves to diffract. In this problem, you will see how different wavelengths diffract as they pass through a city and relate this to reception for radios and cell phones. You will use the angle from the center of the central intensity maximum to the first intensity minimum as a measure of the width of the central maximum (where nearly all of the diffracted energy is found).a. Find the angle θ to the first minimum from thecenter of the central maximum (Express your answer in terms λ and a.):b. What is the angle θFM to the first minimum foran FM radio station with a frequency of 101mMHz? (Express your answer numerically indegrees to three significant figures. Note: Do not write youranswer in terms of trignometric functions. Evaluate any suchfunctions in your working.)c. What is the angle θcell for a cellular phonethat uses radiowaves with a frequency of 900MHz? (Express your answer indegrees to three significant figures.)d. What problem do you encounter in tryingto find the angle θAM for an AM radio stationwith frequency 1000kHz?i. The angle becomes zero.ii. The angle can be given only in radians.iii. To find the angle it would be necessary to takethe arcsine of a negative number.iv. To find the angle it would be necessary totake the arcsine of a number greater than one.
Final answer:
The diffraction of radio waves when passing through city buildings can be represented by single-slit diffraction. The angle to the first minimum can be calculated using the equation θ = sin-1(λ/a), but for AM radio with very large wavelengths, this calculation may not be valid as it could require taking the arcsine of a number greater than one.
Explanation:
When radio waves encounter thin vertical slits such as the spaces between buildings, they diffraction occurs. The property of diffraction can be analyzed using the concept of single-slit diffraction from wave optics. For a single-slit diffraction, the angle θ to the first minimum can be found using the equation θ = sin-1(λ/a), where λ is the wavelength of the wave and a is the width of the slit.
For an FM radio station with a frequency of 101 MHz, we would use the relationship between frequency (f), wavelength (λ), and the speed of light (c) to find the wavelength (λ = c/f) before calculating the angle using the aforementioned equation.
Similarly, for a cellular phone using radio waves with a frequency of 900 MHz, we again find the wavelength using the same relation and then calculate the angle θ to the first minimum.
However, for AM radio, the complication arises because the wavelengths for AM radio are considerably larger. This can lead to a scenario where the slit width is not narrow enough compared to the wavelength, and as a result, the angle θAM calculated using sin-1(λ/a) may result in taking the arcsine of a number greater than one, which is not possible and indicates that the first minimum may not occur.
Suppose that you release a small ball from rest at a depth of 0.600 m below the surface in a pool of water. If the density of the ball is 0.300 that of water and if the drag force on the ball from the water is negligible, how high above the water surface will the ball shoot as it emerges from the water? Neglect any transfer of energy to the splashing and waves produced by the emerging ball.
Answer:
1.4m
Explanation:
You are in a submarine and are at the surface of the ocean but out in the deep sea. There is a big storm and you want to dive down deep enough so that you avoid the turbulence of the sea. You notice the distance between successive waves (wave length) is 20 meters. How far down do you need to dive down to not feel the effect of the waves?
One of the maritime principles that relate the turbulence and wavelength of the waves is called the "depth of 1/2 wavelength" which is also usually referred to as the floor of the wave: A point of depth in which There is no movement. There if a submarine is found, it can be unbalanced and steadily navigate.
If the wavelength is 20 meters, then it must be submerged 10 meters (20/2) to avoid turbulence.
2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane- strain fracture toughness of 98.9 MPa √ m and a yield strength of 860 MPa . The flaw size resolution limit of the flaw de tection appa ratus is 3.0 mm . If the design stress is one- half the yield strength and the value of Y is 1.0, determine whether a critical flaw for this plate is subject to detection.
Answer:
the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Explanation:
This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.
[tex]ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\ ac=0.0168m\\ac=16.8mm[/tex]
Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.
Giraffe bending to drink. In a giraffe with its head 1.83 m above its heart, and its heart 2.04 m above its feet, the(hydrostatic) gauge pressure in the blood at its heart is 246 torr. Assume that the giraffe stands upright and the blood density is 1.06 × 103 kg/m3. In torr (or mm Hg), find the (gauge) blood pressure.
(a) at the brain (the pressure is enough to perfuse the brain with blood, to keep the giraffe from fainting)
(b) at the feet (the pressure must be countered by tight-fitting skin acting like a pressure stocking).
(c) If the giraffe were to lower its head to drink from a pond without splaying its legs and moving slowly, what would be the increase in the blood pressure in the brain? (Such action would probably be lethal.)
Answer:
1) Pm₂ = 1.9 10⁴ Pa , b) P_feet = 5.4 10⁴ Pa , c) Pm₄ = 4.4 10⁴ Pa
Explanation:
a) Pressure can be found using Bernoulli's equation
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rgo g y₂
The amount of blood that runs through the constant system, all the blood that reaches the brain leaves it, so we can assume that the speed of entry and exit of the total blood is the same. In this case the equation is
P₁-P₂ = rgo h (y₂-y₁)
The gauge pressure is
Pm = P₁ -P₂
Pm₂ = 1.06 10³ 9.8 1.83
Pm₂ = 19 10³ Pa
Pm₂ = 1.9 10⁴ Pa
The pressure in the heart is
Pm₁ = 246 torr (1,013 10⁵ Pa / 760 torr) = 3,279 10⁴ Pa
Therefore the gauge pressure is an order of magnitude less
Total or absolute pressure is
Pm₂ = P_heart - P_brain
P_brain = P_heart - Pm₂
P brain = 3,279 10⁴ - 1.9 10⁴
P brain = 1.4 104 Pa
b) on the feet
Pm₃ = rho g y₃
y = 2.04 m
Pm₃ = 1.06 10³ 9.8 2.04
Pm₃ = 21 10³ Pa
Pm₃ = 2.1 10⁴ Pa
Total pressure
Pm₃ = P_feet + P_heart
P_feet = Pm₃ + P_heart
P_feet = 3,279 10⁴ + 2.1 10⁴
P_feet = 5.4 10⁴ Pa
c) If you lower your head the height change is
h = 1.83 +2.04
h = 4.23 m
Pm₄ = 1.06 10³ 9.8 4.23
Pm₄ = 4.4 10⁴ Pa
Two long, parallel wires are attracted to each other by a force per unit length of 350 µN/m. One wire carries a current of 22.5 A to the right and is located along the line y = 0.420 m. The second wire lies along the x axis. Determine the value of y for the line in the plane of the two wires along which the total magnetic field is zero.
Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m
[tex]\dfrac{F}{L}= \dfrac{KI_1I_2}{d}[/tex]
[tex]I_2 = \dfrac{F}{L}\dfrac{d}{KI_1}[/tex]
[tex]I_2 = 350\times 10^{-6}\times \dfrac{0.42}{2 \times 10^{-7}\times 22.5}[/tex]
I₂ = 32.67 A
distance where the magnetic field is zero
[tex]\dfrac{4\pi \times 10^{-7}\times 32.67}{2\pi y_1}=\dfrac{4\pi \times 10^{-7}\times 22.5}{2\pi (0.42-y_1)}[/tex]
[tex]y_1 = 0.248\ m[/tex]
there the distance at which the magnetic field is zero in the two wire is at 0.248 m.
What is the difference between the states of phase equilibrium and metastability?
Answer:
The main difference is that the metastable state is not a state of equilibrium, but a state of non-equilibrium that is maintained for a long time. A metastable state is when the system approaches equilibrium in a very slow manner.
And on the other hand the phase equilibrium as the name says is a state of equilibrium in which there are more than two phases coexisting.
The state of phase is description about the substance existing in equilibrium and metastability is non-equilibrium state of substance.
The given problem is based on the major difference between the states of phase of equilibrium and the metastability. These two are the concepts of chemical equilibrium, when there is subsequent change in the phase of one substance, with respect to the other substance.
The metastable state is not a state of equilibrium, but a state of non-equilibrium that is maintained for a long time. A metastable state is when the system approaches equilibrium in a very slow manner.And on the other hand the phase equilibrium as the name says is a state of equilibrium in which there are more than two phases coexisting.
Thus, we can conclude that the state of phase is description about the substance existing in equilibrium and metastability is non-equilibrium state of substance.
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Select the statement that correctly completes the description of phase difference.
Phase difference describes:
O the difference in the phase angle between any two waves at any given position along the waves.
O the shift between the positions of corresponding crests of two waves of the same frequency.
O the difference in the frequencies of two waves at a given time.
O the displacement of a wave particle from its undisturbed position at the origin.
Phase difference denotes the difference in phase angle between two waves at a given point, occurring when waves are separated by a whole number of multiples of wavelengths.
Explanation:Phase difference describes the difference in the phase angle between any two waves at any given position along the waves. When the waves have the same frequency and the difference in their path lengths is an integer multiple of the wavelength, the waves are said to be in phase. This means these points are separated by a whole number multiple of whole wave cycles or wavelengths. For example, sound waves can illustrate a phase shift when they have different path lengths. It is also important to understand that the wavelength is defined as the distance between any two adjacent points that are in phase.
Air is compressed adiabatically in a piston-cylinder assembly from 1 bar, 300 K to 10 bar, 600 K. The air can be modeled as an ideal gas and kinetic and potential energy effects are negligible. Determine the amount of entropy produced, in kJ/K per kg of air, for the compression. What is the minimum theoretical work input, in kJ per kg of air, for an adiabatic compression from the given initial state to a final pressure of 10 bar?
Final answer:
The entropy production for a reversible adiabatic process of an ideal gas is zero. For the actual work input required for adiabatic compression, the formula involving pressures, volumes, and the heat capacity ratio is used, but cannot be calculated without additional information such as the specific volumes.
Explanation:
To determine the amount of entropy produced during an adiabatic compression of air modeled as an ideal gas, we need to recognize that, by definition, an adiabatic process is one in which no heat is transferred to or from the gas. Therefore, assuming the process is also reversible (which it must be, if we are to calculate a non-zero entropy production), the change in entropy (ΔS) for the process would actually be zero. However, as the conditions stated a rise in temperature during the compression, if any irreversibility were present in the real-world scenario, it would indeed generate entropy, but we need more information to calculate the precise amount for a real-world irreversible process.
The minimum theoretical work input for an adiabatic compression can be calculated using the first law of thermodynamics and the relation for adiabatic processes defined as PV^{γ} = constant, where γ (gamma) is the heat capacity ratio (Cp/Cv). For an ideal gas, the work done (W) on the air during adiabatic compression can be expressed using the formula W = (P2*V2 - P1*V1) / (γ - 1), where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume, respectively. However, since the volume is not given in the question, this calculation cannot be completed without further information.
The entropy produced per kg of air during the adiabatic compression is approximately [tex]\( 0.0348 \)[/tex] kJ/K, and the minimum theoretical work input for the compression is approximately [tex]\( 215.25 \)[/tex] kJ/kg of air.
The amount of entropy produced during the adiabatic compression process in a piston-cylinder assembly, where air is modeled as an ideal gas, can be determined using the following relationship for an adiabatic process:
[tex]\[ s_2 - s_1 = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right) \][/tex]
where [tex]\( s \)[/tex] is the specific entropy, [tex]\( c_p \)[/tex] is the specific heat at constant pressure, [tex]\( T \)[/tex] is the absolute temperature, [tex]\( R \)[/tex] is the specific gas constant, and [tex]\( P \)[/tex] is the absolute pressure. The subscripts [tex]\( 1 \) and \( 2 \)[/tex] denote the initial and final states, respectively.
Given:
- Initial state: [tex]\( P_1 = 1 \) bar, \( T_1 = 300 \)[/tex] K
- Final state: [tex]\( P_2 = 10 \) bar, \( T_2 = 600 \)[/tex] K
- For air, [tex]\( c_p = 1.005 \) kJ/kg\K and \( R = 0.287 \) kJ/kg[/tex]
First, convert the pressures from bar to Pa (since the gas constant [tex]\( R \)[/tex] is in terms of Pa):
- [tex]\( P_1 = 1 \times 10^5 \) Pa[/tex]
- [tex]\( P_2 = 10 \times 10^5 \) Pa[/tex]
Now, calculate the entropy change:
[tex]\[ s_2 - s_1 = 1.005 \ln\left(\frac{600}{300}\right) - 0.287 \ln\left(\frac{10 \times 10^5}{1 \times 10^5}\right) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \ln(2) - 0.287 \ln(10) \][/tex]
[tex]\[ s_2 - s_1 = 1.005 \times 0.693 - 0.287 \times 2.303 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.6965 - 0.6617 \][/tex]
[tex]\[ s_2 - s_1 \approx 0.0348 \text{ kJ/kg\K} \][/tex]
For the minimum theoretical work input during an adiabatic compression, we use the following equation for an ideal gas:
[tex]\[ w_{\text{in,min}} = \int_{V_1}^{V_2} P \, dV \][/tex]
For an adiabatic process, [tex]\( PV^\gamma = \text{constant} \), where \( \gamma \)[/tex] is the heat capacity ratio ([tex]\( c_p/c_v \)[/tex]). The work input can be calculated as:
[tex]\[ w_{\text{in,min}} = \frac{P_2V_2 - P_1V_1}{\gamma - 1} \][/tex]
Using the ideal gas law, [tex]\( PV = mRT \)[/tex], where [tex]\( m \)[/tex] is the mass of the gas, we can express [tex]\( V \)[/tex] in terms of [tex]\( P \) and \( T \)[/tex]:
[tex]\[ V = \frac{mRT}{P} \][/tex]
Since the mass [tex]\( m \)[/tex] cancels out, we can write:
[tex]\[ w_{\text{in,min}} = \frac{mRT_2 - mRT_1}{\gamma - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{R(T_2 - T_1)}{\gamma - 1} \][/tex]
Given that [tex]\( \gamma = \frac{c_p}{c_v} = \frac{c_p}{c_p - R} \)[/tex], we can calculate \( \gamma \) for air:
[tex]\[ \gamma = \frac{1.005}{1.005 - 0.287} \approx 1.4 \][/tex]
Now, calculate the minimum work input:
[tex]\[ w_{\text{in,min}} = \frac{0.287(600 - 300)}{1.4 - 1} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{0.287 \times 300}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} = \frac{86.1}{0.4} \][/tex]
[tex]\[ w_{\text{in,min}} \approx 215.25 \text{ kJ/kg} \][/tex]
The period of a sinusoidal source is the time required for the sinusoid to pass through all of its possible values. We use the symbol T to represent the period of a sinusoid. The period and the frequency are inversely related. A sinusoidal source described by the function cos(ωt) has a frequency of ω radians/second, or a frequency f=ω/2π Hz. The units hertz represents the number of cycles per second. Since the period is the number of seconds per cycle, the period is the inverse of the frequency in hertz: T=1f Substituting the frequency in radians/second, ω, for the frequency in Hz gives us another way to calculate the period: T=2πω What is the period of the voltage source described as v(t)=50cos(2000t−45∘) mV? Express your answer to two digits after the decimal point and include the appropriate units.
Answer:
T=0.0031secs
Explanation:
The voltage expression [tex]v(t)=50cos(2000t-45^{0})[/tex] can be represented as
[tex]v(t)=v_{m}cos(wt-\alpha ) \\[/tex]
comparing the two equations we can conclude that the angular frequency
[tex]w=2000[/tex]
from the question, since the frequency,f which is express as
[tex]f=\frac{w}{2\pi }\\[/tex],
Hence [tex]f=\frac{2000}{2\pi } \\f=\frac{2000}{2*3.14 } \\f=318.471Hz\\[/tex].
The period which is the inverse of the frequency can be express as
[tex]T=\frac{1}{f} \\T=\frac{1}{314.471}\\ T=0.00314\\T=0.0031secs[/tex]
A 92kg astronaut and a 1200kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.14m/s directly away from the shuttle. Seven and a half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?
Answer:
13.7m
Explanation:
Since there's no external force acting on the astronaut or the satellite, the momentum must be conserved before and after the push. Since both are at rest before, momentum is 0.
After the push
[tex]m_av_a + m_sv_s = 0[/tex]
Where [tex]m_a = 92kg[/tex] is the mass of the astronaut, [tex]m_s = 1200kg[/tex] is the mass of the satellite, [tex]v_s = 0.14 m/s[/tex] is the speed of the satellite. We can calculate the speed [tex]v_a[/tex] of the astronaut:
[tex]v_a = \frac{-m_sv_s}{m_a} = \frac{-1200*0.14}{92} = -1.83 m/s[/tex]
So the astronaut has a opposite direction with the satellite motion, which is further away from the shuttle. Since it takes 7.5 s for the astronaut to make contact with the shuttle, the distance would be
d = vt = 1.83 * 7.5 = 13.7 m
This is a conservation of momentum problem where the astronaut moves opposite to the direction of the satellite's movement due to Newton's third law. The astronaut's velocity is calculated using the conservation of momentum principle, and the distance between him and the shuttle is then determined via the formula for distance.
Explanation:This problem involves understanding the conservation of momentum in a system with no external forces acting on it. When an astronaut pushes a satellite in space, there's a reaction force acting back on the astronaut due to Newton's third law. So, the astronaut will also move in the opposite direction. Keep in mind that the net momentum before and after this action remains zero as there are no external forces.
We're given that the astronaut comes into contact with the shuttle seven and half seconds after pushing on the satellite. He must have been moving at a certain speed to cover the distance in this time. Due to conservation of momentum, we can set up an equation as follows: Momentum of Astronaut + Momentum of Satellite = 0 (Because initially they were at rest). We can then calculate this to find the velocity of the astronaut.
After getting the velocity of the astronaut, we use the formula for distance: Distance = Speed * Time to get the initial distance between astronaut and Shuttle.
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A grinding wheel is a uniform cylinder with a radius of 7.80 cm and a mass of 0.550 kg.
Part A
Calculate its moment of inertia about its center. Express your answer to three significant figures and include the appropriate units.
Part B
Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 7.40 s .
Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 58.0 s .
Answer:
a. I = 167.31 x 10 ⁻³ kg*m²
b. T = 4.59 kg * m² / s²
Explanation:
The moment of inertia of a uniform cylinder:
a.
r = 7.8 cm * 1 m / 100 cm = 0.078 m
I = ½ * m * r²
I = ½ * 0.55 kg * (0.078²m)
I = 167.31 x 10 ⁻³ kg*m²
b.
T = Iα’ + Iα,
α’ = ω’/t = 1750 rpm * (2π/60) / 7.40s = 24.76 rad/s²
α = ω/t = 1500 rpm * (2π/60) / 58 = 2.71 rad/s²
T = (167.31 x 10⁻³ kg*m²)* (24.76 + 2.71 ) rad / s²
T = 4.59 kg * m² / s²
The moment of inertia of the wheel is calculated as 0.00133 kg*m^2. The second part of the question involves determining the net and frictional torques to find the total applied torque.
Explanation:To solve this problem, we need to apply the formulas of moment of inertia and the angular acceleration along with the concept of frictional torque. The moment of inertia for a cylinder rotating about its axis is given by the formula I = 0.5*m*r^2. In this case, where mass (m) is 0.550 kg and radius (r) is 7.80 cm or 0.078 m (since 1cm = 0.01m).
Part A: I = 0.5 * 0.550 kg * (0.078 m)^2 = 0.00133 kg*m^2.
For Part B, we first need to convert the rotational speed from revolutions per minute (rpm) to rad/s. Then we use these values to determine the angular acceleration and calculate the net torque. The frictional torque is then added to this net torque to find the total applied torque.
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In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg, roll toward each other and collide. The velocity is measured before and after each collision. The collected data is shown below. A 5 column table with 3 rows. The first column is unlabeled with entries Trial 1, Trial 2, Trial 3, Trial 4. The second column is labeled Initial Velocity Ball A (meters per second) with entries positive 1, positive 0.5, positive 2, positive 0.5. The third column is labeled Initial Velocity Ball B (meters per second) with entries negative 2, negative 1.5, positive 1, negative 1. The third column is labeled Final Velocity Ball A (meters per second) with entries negative 2, negative 0.5, positive 1, positive 1.5. The fourth column is labeled Final Velocity Ball B (meters per second) with entries negative 1, negative 0.5, negative 2, negative 1.5. Which trial shows the conservation of momentum in a closed system? Trial 1 Trial 2 Trial 3 Trial 4
Answer:
Second Trial satisfy principle of conservation of momentum
Explanation:
Given mass of ball A and ball B [tex]=\ 1.0\ Kg.[/tex]
Let mass of ball [tex]A[/tex] and [tex]B\ is\ m[/tex]
Final velocity of ball [tex]A\ is\ v_1[/tex]
Final velocity of ball [tex]B\ is\ v_2[/tex]
initial velocity of ball [tex]A\ is\ u_1[/tex]
Initial velocity of ball [tex]B\ is\ u_2[/tex]
Momentum after collision [tex]=mv_1+mv_2[/tex]
Momentum before collision [tex]= mu_1+mu_2[/tex]
Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.
Now, [tex]mu_1+mu_2=mv_1+mv_2[/tex]
Plugging each trial in this equation we get,
First Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
Second Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1[/tex]
moment before collision [tex]=[/tex] moment after collision
Third Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
Fourth Trial
[tex]mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0[/tex]
momentum before collision [tex]\neq[/tex] moment after collision
We can see only Trial- 2 shows the conservation of momentum in a closed system.
Answer: Trial 2
Explanation:
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K. You set the bowl up so that as it cools to room temperature the heat first flows through a Carnot Engine. The soup has Cv= (33 J/K). Assume that the volume of the soup does not change.
1. What fraction of the total heat QH that is lost by the soup can be turned into useable work by the engine?
Answer:
Explanation:
Heat energy given out by the soup
= C_v x ( t₂ - t₁ )
= 33 x ( 340 - 300)
= 1320 J
This heat is given to Carnot engine . Efficiency of engine
= (340 - 300 ) / 340
= 40 / 340
2 / 17
This fraction of total heat given is converted into useable work by the engine.
A 90.0-kg fullback running east with a speed of 5.00 m/s is tackled by a 95.0-kg opponent running north with a speed of 3.00 m/s. (a) Why does the tackle constitute a perfectly inelastic collision? (b) Calculate the velocity of the players immediately after the tackle and (c) determine the mechanical energy that is lost as a result of the collision. (d) Where did the lost energy go?
Answer:
a) Please see below as the answer is self-explanatory.
b) 2.88 m/s
c) 785. 8 J
d) It is expended like thermal energy, due to internal friction.
Explanation:
a) In a tackle, both players keep emmeshed each other, so it is a perfectly inelastic collision; Immediately after the tackle, both masses behave like they were only one.
b) Assuming no external forces act during the collision, total momentum must be conserved.
As momentum is a vector, the conservation principle must be met by all vector components at the same time.
In our case, as the players move in directions mutually perpendicular, we can decompose the momentum vector along both directions, taking into account that after the collision, the momentum vector will have components along both directions.
So, if we call the W-E axis our X-axis (being the direction towards east as the positive one) , and to the S-N axis our Y -axis (being the northward direction the positive one), we can write the following equations:
pₓ₀ = pₓf ⇒ m₁*v₁ = (m₁+m₂)*vf*cosθ
py₀ = pyf ⇒ m₂*v₂ = (m₁+m₂)*vf*sin θ
where θ, is the angle that both players take regarding the x-axis after the collision (north of east).
Replacing by the values, we have the following equations:
vf*cosθ = (90.0 kg*5.00 m/s) / (90.0 kg + 95.0 kg) = 2.43 m/s (1)
vf*sin θ = (95.0 kg* 3.00 m/s) / (90.0 kg + 95.0 kg) = 1.54 m/s (2)
Dividing both sides:
sin θ / cos θ = tan θ = 1.54 / 2.43 = 0.634
⇒ arc tan (0.634) = 32.3º
Replacing in (1) we have:
vf = 2.43 m/s / cos 32.3º = 2.43 m/s / 0.845 = 2.88 m/s
c) As the collision happens in one dimension, all mechanical energy, before and after the collision, is just the kinetic energy of the players.
Before the collision:
K₀ = 1/2*m₁*v₁₀² + 1/2 m₂*v₂₀²
= 1/2*( ( 90.0) kg*(5.0)²(m/s)² + (95.0)kg*(3.0)(m/s)²) = 1,553 J
After the collision:
Kf = 1/2 *(m₁+ 767.2 Jm₂)*vf² = 1/2*185 kg*(2.88)²(m/s)²= 767.2 J
The mechanical energy lost during the collision is just the difference between the final and initial kinetic energy:
ΔK = Kf - K₀ = 767.2 - 1,553 J = -785.8 J
So, the magnitude of the energy lost during the collision is 785.8 J.
d) This energy is lost during the collision as thermal energy, due to the internal friction between both players.
The tackle constitutes a perfectly inelastic collision where the players stick together after the collision, resulting in a loss of kinetic energy. The velocity of the players immediately after the tackle is 2.70 m/s to the east. The mechanical energy lost as a result of the collision is 562.5 J.
Explanation:(a) The tackle constitutes a perfectly inelastic collision because the two players stick together after the collision, resulting in a loss of kinetic energy. In a perfectly inelastic collision, the objects involved stick together and move as a single unit.
(b) To calculate the velocity of the players immediately after the tackle, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. Since the fullback is running east, we can consider the positive direction as east and the negative direction as north. Applying the principle of conservation of momentum in the x-direction, we have:
Total momentum before the collision in the x-direction: (90.0 kg)(5.00 m/s) = 450 kg·m/sTotal momentum after the collision in the x-direction: (90.0 kg + 95.0 kg) * Vx = (185.0 kg) * VxSetting the two equations equal to each other and solving for Vx, we get:
(90.0 kg)(5.00 m/s) = (185.0 kg) * VxVx = 2.70 m/sSo the velocity of the players immediately after the tackle is 2.70 m/s to the east.
(c) The mechanical energy that is lost as a result of the collision can be calculated by subtracting the final kinetic energy from the initial kinetic energy. The initial kinetic energy is given by:
Initial kinetic energy = 0.5 * (90.0 kg) * (5.00 m/s)^2 = 562.5 JSince the players come to rest after the collision, the final kinetic energy is zero. Therefore, the mechanical energy lost is equal to the initial kinetic energy:
Mechanical energy lost = Initial kinetic energy = 562.5 J(d) The lost energy is converted into other forms of energy, such as sound, heat, and deformation of the players and their surroundings.
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A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A constant tangential force of 210 N applied to its edge causes the wheel to have an angular acceleration of 0.932 rad/s2. What is the moment of inertia of the wheel? (Pick the answer closest to the true value.)A. 27.3 kg m2B. 42.4 kg m2C. 54.9 kg m2D. 74.4 kg m2E. 98.5 kg m2
Answer:
Moment of inertia will be [tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
Explanation:
We have given radius of solid cylinder r = 0.330 m
Constant tangential force F = 210 N
Angular acceleration [tex]\alpha =0.932rad/sec^2[/tex]
We know that torque [tex]\tau =Fr=210\times 0.330=69.3Nm[/tex]
We also know that torque is given by [tex]\tau =I\alpha[/tex]
So [tex]69.3=I\times 0.932[/tex]
[tex]I=74.356kgm^2[/tex]
So option (d) will be the correct answer
If the coefficient of static friction between tires and pavement is 0.60, calculate the minimum torque that must be applied to the 69-cm-diameter tire of a 920-kg automobile in order to "lay rubber" (make the wheels spin, slipping as the car accelerates). Assume each wheel supports an equal share of the weight.
To solve this problem, it is necessary to apply the definitions and concepts related to Newton's second law, which relate the variables of the Normal Force, Weight, friction force and finally the Torque.
We start under the definition that the Normal Force of one of the 4 tires of the car would be subject to
[tex]N = \frac{mg}{4}[/tex]
Where,
m = mass
g = Gravitational Acceleration
Therefore the Normal Force of each wheel would be
[tex]N = \frac{920*9.8}{4}[/tex]
[tex]N = 2254N[/tex]
Now the friction force can be determined as
[tex]f_s = \mu_s N[/tex]
[tex]f_s = 0.60 * 2254[/tex]
[tex]f_s = 1352.4N[/tex]
The radius of each of the tires is given as
[tex]r = \frac{69}{2}[/tex]
[tex]r = 34.5cm = 0.345m[/tex]
Finally, the torque is made between the friction force (which is to be overcome) and the radius of each of the wheels, therefore:
[tex]\tau = r*f_s[/tex]
[tex]\tau = (0.345)(1352.4)[/tex]
[tex]\tau = 466.578N\cdot m[/tex]
Therefore the engine of the car must apply a torque of about [tex]466.578N\cdot m[/tex] to lay rubber
To calculate the minimum torque needed to make the wheels spin on a car, we must first understand the concept of static friction. The force of static friction (Fs) that must be overcome to cause slipping is given by
Fs = μsN
where μs is the coefficient of static friction and N is the normal force. In this case, the weight of the car (W) is evenly distributed on all four tires, so each tire supports a quarter of the weight, W/4. The normal force N for one tire would then be W/4.
Since the weight W of the car is the mass (m) times the acceleration due to gravity (g), we have:
N = W/4 = mg/4.
Substituting the given values, we find
N = (920 kg * 9.81 m/s2)/4.
Using the coefficient of static friction (μs = 0.60), the static frictional force Fs for one tire is
Fs = 0.60 * N.
To find the torque (τ), we use the relation
τ = Fsr
where r is the radius of the tire.
The radius is half the diameter, so r = 69 cm / 2 or 0.345 m. Thus, the minimum torque is
τ = Fs * 0.345 m.
Calculating N, we get
N = (920 kg * 9.81 m/s2)/4
N = 2251.05 N
so Fs = 0.60 * 2251.05 N
Fs = 1350.63 N.
Therefore, the minimum torque τ is 1350.63 N * 0.345 m = 465.97 Nm.
The St. Louis Arch has a height of 192 m. Suppose that a stunt woman of mass 84 kg jumps off the top of the arch with an elastic band attached to her feet. She reaches the ground at zero speed. The acceleration of gravity is 9.81 m/s 2 . Find her kinetic energy after 2.6 s of the flight. Assume the elastic band has no length and obeys Hooke’s Law. Answer in units of kJ
Final answer:
The kinetic energy of the stunt woman after 2.6 s of flight is 27578.835 kJ.
Explanation:
To find the kinetic energy of the stunt woman after 2.6 s of flight, we can use the formula for kinetic energy:
KE = 0.5 mv^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
First, let's find the velocity of the stunt woman after 2.6 s of flight. We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity (0 m/s), a is the acceleration due to gravity (-9.81 m/s^2), and t is the time (2.6 s).
Substituting the values, we get:
v = 0 + (-9.81) * 2.6
v = -25.446 m/s
Since the stunt woman reaches the ground at zero speed, her final velocity is 0 m/s. Therefore, her velocity is -25.446 m/s after 2.6 s of flight.
Now, let's plug the values of mass (84 kg) and velocity (-25.446 m/s) into the formula for kinetic energy:
KE = 0.5 * 84 * (-25.446)^2
KE = 0.5 * 84 * 650.701716
KE = 27578.835 kJ
Therefore, the stunt woman's kinetic energy after 2.6 s of flight is 27578.835 kJ.
The intensity level of a power mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are mowing their lawn 20 m from your open bedroom window. What is the intensity level in your bedroom? The intensity level of a power mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that four of your neighbors are mowing their lawn 20 m from your open bedroom window. What is the intensity level in your bedroom?
A. 50 dB
B. 400 dB
C. 104 dB
D. 80 dB
E. 40 dB
Answer:
β₂ = 74 dB, The answer is D which is the closest
Explanation:
The definition and intensity is the power per unit area
I = P / A
P = I A
The emitted power is constant whereby the energy is distributed over the surface of a sphere
A = 4π R²
We can also write it in two points
P = I₁ A₁ = I₂ A₂
I₁ / I₂ = A₂ / A₁
I₁ / I₂ = 4π R₂² / 4π R₁²
I₁ / I₂ = R₂² / R₁²
The definition of decibels is
β = 10 log (I / I₀)
Let's write this equation for the two given points
m = 1m
β₁ = 10 log (I₁ / I₀)
m = 20m
β₂ = 10 log (I₂ / I₀)
Let's eliminate I₀
β₁ - β₂ = 10 log (I₁ / I₀) - 10 log (I₂ / I₀) = 10 (log (I₁ / I₀) –log (I₂ / I₀))
β₁ - β₂ = 10 log (I₁ / I₂)
β₁ - β₂ = 10 log (R₂² / R₁²)
Let's calculate
100 –β₂ = 10 log (20²/1²)
β₂ = 100 - 10 log 400
β₂ = 100 - 26.0
β₂ = 74 dB
The answer is D which is the closest
The intensity level in the bedroom is approximately 74 dB.
Explanation:The intensity level of sound decreases as the distance from the source increases. With each doubling of distance, the sound intensity decreases by 6 dB. In this case, the sound mowers are at a distance of 20 m from your bedroom window, which is 20 times the distance of 1.0 m where the intensity level is 100 dB. Therefore, the intensity level in your bedroom would be 100 dB - (6 dB x log2(20)) = 100 dB - 6 dB x 4.32 = 100 dB - 25.92 dB ≈ 74 dB.
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Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements that will allow you to determine the charge on each sphere. You set up a coordinate system with one sphere (charge q1) at the origin and the other sphere (charge q2) at x = +0.400 m. Available to you are a third sphere with net charge q3 = 3.00×10−6 C and an apparatus that can accurately measure the location of this sphere and the net force on it. First you place the third sphere on the x-axis at x = 0.200 m; you measure the net force on it to be 6.50 N in the +x-direction. Then you move the third sphere to x = +0.600 m and measure the net force on it now to be 3.50 N in the +x-direction.
Part A) Calculate q1.
Part B) Calculate q2.
Answer:
a) q₁ = 15. 28 10⁻⁶C, b) q₂ = 5.64 10⁻⁶ C
Explanation:
For this exercise we use Newton's second law where force is Coulomb's electric force
Case 1. Distance (x₁ = 0.200 m) from the third sphere
F₁ = F₁₃ - F₂₃
F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²
F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²
Case2 Distance (x₂ = 0.6 m) from the third sphere
F₂ = F₁₃ + F₂₃
F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²
F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²
The distance is between the spheres, in the annex you can see the configuration of the charge and forces
Let's replace the values
F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²
F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²
6.50 = 674. 25 10³ (q₁ –q₂)
3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)
We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second
q₁ = q₂ + 6.50 / 674 10³
3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04
1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶
q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶
q₂ 27,777 = 156,557 10⁻⁶
q₂ = 156.557 10-6 /27.777
q₂ = 5.636 10⁻⁶ C
We look for q1 in the other equation
q₁ = q₂ + 6.50 / 674 10³
q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶
q₁ = 15. 28 10⁻⁶C
Final answer:
To find the charges on the two spheres, we can use Coulomb's law. Calculations show that q1 is approximately 4.333 × 10^-8 C and q2 is approximately 1.111 × 10^-7 C.
Explanation:
To determine the charges on the two spheres, we can use Coulomb's law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Let's calculate the charges:
Part A) Calculating q1:
Using the information given, we can set up the following equation:
F1 = k * (q1 * q3) / (0.200)², where F1 is the net force at x = 0.200 m.
Substituting the given values, we get:
6.50 = (9 × 10^9) * (q1 * (3 × 10^-6)) / (0.200)².
Simplifying the equation, we find that q1 = 4.333 × 10^-8 C.
Part B) Calculating q2:
Using the same equation, but with F2, the net force at x = 0.600 m, we have:
3.50 = (9 × 10^9) * (q2 * (3 × 10^-6)) / (0.600)².
Simplifying, we find that q2 = 1.111 × 10^-7 C.#SPJ11
What is the difference between transverse and longitudinal waves?
a. Only longitudinal waves transmit matter.
b. Plane waves are transverse waves while spherical waves are longitudinal.
c. Only transverse waves transmit energy.
d. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.
e. Mechanical waves are transverse waves while electromagnetic waves are longitudinal.
To explain how transverse and longitudinal waves work, let us give two examples for each particular case.
In the case of transverse waves, the displacement of the medium is PERPENDICULAR to the direction of the wave. One way to visualize this effect is when you have a rope and between two people the rope is shaken horizontally. The shift is done from top to bottom. This phenomenon is common to see it in solids but rarely in liquids and gases. A common application usually occurs in electromagnetic radiation.
On the other hand in the longitudinal waves the displacement of the medium is PARALLEL to the direction of propagation of the wave. A clear example of this phenomenon is when a Slinky is pushed along a table where each of the rings will also move. From practice, sound waves enclose the definition of longitudinal wave displacement.
Therefore the correct answer is:
C. In transverse waves the displacement is perpendicular to the direction of propagation of the wave, while in longitudinal waves the displacement is parallel to the direction of propagation.
Transverse waves have a perpendicular disturbance while longitudinal waves have a parallel disturbance.
Explanation:A transverse wave has a disturbance perpendicular to its direction of propagation, whereas a longitudinal wave has a disturbance parallel to its direction of propagation. For example, when you ripple a string up and down, you create a transverse wave. But when you push and pull a slinky back and forth, you create a longitudinal wave.
While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 17m , and you use your watch to find that each loop around takes 26sWhat is your speed?Express your answer to two significant figures and include the appropriate units.What is the magnitude of your acceleration?Express your answer to two significant figures and include the appropriate units.What is the ratio of your weight at the top of the ride to your weight while standing on the ground?Express your answer using two significant figures.
Answer:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Explanation:
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
Case Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed theCase Study: General Andrew Jackson: Andrew Jackson's military career spanned several wars including the American Revolution, the Creek War, the War of 1812, and the First Seminole War. After the Creek War, Jackson and the Creek Indians signed the
A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m per meter when the shaft rotates. The shaft is to be used for applying a torque of 0.1 N m to actuate a switch. What is the maximum length of shaft that may be used if the shear stress in the shaft is not to exceed 50 MPa?
Answer:
2.5m
Explanation:
Torque is defined as the rotational effect of a force on a body.
The torque T for the maximum shear stress is given as 0.1 Nm
Frictional torque is the torque caused by a frictional force
The frictional torque F is given as 0.04 Nm/m
The maximum length of the shaft is thus given as
L = T / F
= 0.1/0.04
L= 2.5 m
The mean diameters of Mars and Earth are 6.9 x 10^3 km and 1.3 x 10^4 km, respectively. The mass of Mars is 0.11 times Earth’s mass.
a) What is the ratio of the mean density (mass per unit volume) of Mars to that of Earth?
b) What is the value of the gravitational acceleration on Mars?
c) What is the escape speed on Mars?
Final answer:
a) The ratio of the mean density of Mars to that of Earth is 0.11 times the ratio of the volume of Earth to the volume of Mars. b) The value of the gravitational acceleration on Mars is 3.7 m/s². c) The escape speed on Mars is 5.03 km/s.
Explanation:
a) To find the ratio of the mean density of Mars to that of Earth, we need to divide the mass of Mars by the volume of Mars and divide the mass of Earth by the volume of Earth. The mean density is given by:
Mean density (Mars) = mass (Mars) / volume (Mars)
Mean density (Earth) = mass (Earth) / volume (Earth)
Substituting the given values, we have:
Mean density (Mars) = (0.11 x mass (Earth)) / volume (Mars)
Mean density (Earth) = mass (Earth) / volume (Earth)
Dividing these two equations, we get the ratio of the mean densities as:
Ratio of mean density (Mars to Earth) = (0.11 x mass (Earth)) / volume (Mars) / (mass (Earth) / volume (Earth))
Simplifying, the ratio of mean densities is 0.11 times the ratio of the volume of Earth to the volume of Mars.
b) The value of the gravitational acceleration on Mars can be found using Newton's law of gravitation. The formula for gravitational acceleration is:
Gravitational acceleration = (Gravitational constant * mass of Mars) / radius of Mars^2
Substituting the given values, we have:
Gravitational acceleration on Mars = (6.67 x 10^-11 N m^2/kg^2 * 6.418 x 10^23 kg) / (3.38 x 10^6 m)^2 = 3.7 m/s^2
c) The escape speed on Mars can be found using the formula:
Escape speed = sqrt(2 x Gravitational constant x mass of Mars / radius of Mars)
Substituting the given values, we have:
Escape speed on Mars = sqrt(2 x 6.67 x 10^-11 N m^2/kg^2 x 6.418 x 10^23 kg / 3.38 x 106 m) = 5.03 km/s
A single-turn circular loop of wire of radius 50 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10-s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. (a) Determine the emf induced in the loop. (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
The magnitude of the induced emf in the loop can be calculated using Faraday's law of electromagnetic induction. The direction of the induced current that would oppose the change in flux can be determined using Lenz's law and the right-hand rule, which in this case would be clockwise.
Explanation:To answer your questions: (a) emf induced in the loop and (b) the direction of the current induced in the loop, we must understand Faraday's law of electromagnetic induction and Lenz's law.
Using Faraday's law, the magnitude of the induced emf is given as ε = |døm/dt|, where øm is the magnetic flux.
The magnetic flux through a loop is given by the product of the magnetic field B, the area A (Which can be calculated using the formula for the area of a circle, A = πr², where r is the radius of the loop), and the cosine of the angle Ө between the magnetic field lines and the perpendicular to the plane of the loop.
However, since the loop is perpendicular to the magnetic field, cosӨ = 1. Also, because the magnetic field B is spatially uniform and changes uniformly with time, we can find the average magnetic field over the 0.1 s interval as B = (200mT + 300mT) / 2 = 250mT = 0.25 T. Therefore, the change in magnetic flux over the time interval is Δøm = BA = (0.25 T)(π(0.05 m)²). Substituting these into Faraday's law gives us the induced emf.
According to Lenz's law, the direction of the induced emf (and hence the current) is such that it produces a magnetic field that opposes the change in the magnetic flux. Since the original field is directed out of the page and is increasing, the induced current must flow in a way that it produces a field into the page. Hence, according to the right-hand rule, the current would flow clockwise when viewed from above.
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By applying Faraday's Law of electromagnetic induction, the induced electromotive force (emf) in the loop is calculated to be approximately 7.85 mV. Following Lenz's Law, the induced current flows in a clockwise direction to oppose the increase in the magnetic field.
To address the student's question, we can use Faraday's Law of electromagnetic induction which states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit.
(a) To find the induced emf, use the formula:
E = -dΦ/dt
where Φ is the magnetic flux given by:
Φ = BA cos(θ)
B is the magnetic field's magnitude, A is the area of the loop, and θ is the angle between the field and normal to the loop. In this case, θ is 0 degrees because the field is perpendicular to the plane of the loop.
Φ = B(πr²)
Given that r = 0.05 m, Binitial = 200 mT, Bfinal = 300 mT, and Δt = 0.10 s, we have:
ΔΦ = π(0.05)²(300 mT - 200 mT)
ΔΦ = π(0.05)²(100 mT) = π(0.0025 m²)(100 x 10⁻³ T) = π(0.00025 T·m²)
E = -ΔΦ/Δt = -π(0.00025)/0.10 = -0.00785 V
The magnitude of the induced emf is approximately |0.00785| V or 7.85 mV.
(b) According to Lenz's Law, the direction of the induced current will be such that it opposes the change in flux. Since the magnetic field is increasing and directed out of the page, the induced current will create a magnetic field into the page to oppose the increase. Using the right-hand rule, we can determine that the current flows in a clockwise direction when viewed from above.
Two blocks of ice , one four times as heavy as the other, aren at rest on a frozen lake. a person pushes each block the same distance d.ignore friction and assume that an equal force F is exerted on each block. which of the following statements is true about the kinetic energy of the heavier block after the push?A) it is smaller than kinetic energy of the lighter blockB) it is equal to the kinetic energy of the lighter blockC) it is larger than the kinetic energy of the lighter blockD) it cannot be determined without knowing the force and the mass of each block
Answer:
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Explanation:
Hi there!
The work done on each block is calculated as follows:
W = F · d
Since the two blocks were pushed the same distance with the same force, the work done on each object is the same.
Using the work-energy theorem, we know that the work done on an object is equal to its change in kinetic energy (KE):
W = ΔKE
W = final KE - initial KE
Since the objects are at rest, initial KE = 0, then:
W = final KE
Since the work done on each block is the same, so will be its final kinetic energy.
The correct answer is B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block.
Final answer:
When an equal force is applied over the same distance to two blocks of ice, with one being four times heavier than the other, the kinetic energy gained by both blocks is equal, because the work done on them is the same.
Explanation:
The question involves two blocks of ice on a frozen lake, with one block being four times as heavy as the other. When an equal force F is exerted on each block, pushing them the same distance d, we are asked which statement is true about the kinetic energy of the heavier block after the push compared to the kinetic energy of the lighter block. Since the work done (work = force × distance) on both blocks is the same and work done on an object is equal to the change in its kinetic energy (Work-Energy Principle), both blocks will have the same increase in kinetic energy. Therefore, without considering initial kinetic energies (since both start from rest), the correct answer is that it is equal to the kinetic energy of the lighter block (Option B). This is because the amount of work done on both blocks is the same, leading to an equal increase in kinetic energy.