Answer:
The horizontal force that must be applied to the box to cause it to start sliding along the surface is 162N
Explanation:
To start sliding the box on the surface it must overcome its static frictional force under equilibrium condition
The net force on the box is
F - fs = 0
F = fs = us N = us mg
Force = ( 0.3) x ( 55 kg) x ( 9.8 m/s^2) = 161.7 approximately 162 N
Horizontal force is defined as the forces that equal and opposite in the direction. The horizontal resultant force is always zero.
Given that:
Mass of box = 55 Kg
Coeeficient of Static friction = 0.30
Coefficient of kinetic friction = 0.20
The box if have to slide, then it would have to overcome the static force, such that:
F - Fs = 0
F = Fs
The force will be:
F = [tex]\rm \mu \times mass \times g[/tex]
F = [tex]0.3 \times 55 \times 9.8[/tex]
F = 161.7 Newton
Thus, the force required to slide the box is 161.7 Newton.
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Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart.
Answer:
av=0.333m/s, U=3.3466J
b.
[tex]v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s[/tex]
Explanation:
a. let [tex]m_A[/tex] be the mass of block A, and[tex]m_B=10.0kg[/tex] be the mass of block B. The initial velocity of A,[tex]\rightarrow v_A_1=2.0m/s[/tex]
-The initial momentum =Final momentum since there's no external net forces.
[tex]pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):
[tex]v_A_1-v_B_1=v_{B2}-v_{A2}[/tex]
-Applying the conservation of momentum. The blocks have the same velocity after collision:
[tex]v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s[/tex]
#Total Mechanical energy before and after the elastic collision is equal:
[tex]K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J[/tex]
Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s
b. Taking the end collision:
From a above, [tex]m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0[/tex]
We plug these values in the equation:
[tex]m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]
[tex]2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s[/tex]
(a) The maximum energy stored in the spring bumpers during the collision is 3.00 J, and the velocity of both the blocks is 0.50 m/s.
(b) After they move apart, block A has a velocity of -1.00 m/s and block B has a velocity of 1.00 m/s.
You can follow these simple steps to find the required solution -
(a) Maximum Energy Stored in the Spring Bumpers
To find the maximum energy stored in the spring bumpers, we will use the conservation of momentum and energy principles.
Initially, block A (mass 2.00 kg) is moving at 2.00 m/s, and block B (mass 6.00 kg) is at rest. The total initial momentum ([tex]p_{initial[/tex]) is:
[tex]p_{\text{initial}} = m_A \cdot v_A + m_B \cdot v_B = 2.00 \, \text{kg} \cdot 2.00 \, \text{m/s} + 6.00 \, \text{kg} \cdot 0 \, \text{m/s} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]At the point of maximum compression, both blocks momentarily move with the same velocity ([tex]v_{common[/tex]). Using the conservation of momentum:
[tex]p_{\text{final}} = (m_A + m_B) \cdot v_{\text{common}} = 4.00 \, \text{kg} \cdot \text{m/s} \\[/tex]Solving for [tex]v_{common[/tex]:
[tex]v_{\text{common}} = \frac{4.00 \, \text{kg} \cdot \text{m/s}}{2.00 \, \text{kg} + 6.00 \, \text{kg}} = 0.50 \, \text{m/s} \\[/tex]Next, determine the initial kinetic energy ([tex]KE_{initial[/tex]):
[tex]KE_{\text{initial}} = 0.5 \cdot m_A \cdot (v_A)^2 + 0.5 \cdot m_B \cdot (v_B)^2 = 0.5 \cdot 2.00 \, \text{kg} \cdot (2.00 \, \text{m/s})^2 + 0.5 \cdot 6.00 \, \text{kg} \cdot (0 \, \text{m/s})^2 = 4.00 \, \text{J} \\[/tex]The kinetic energy at the point of maximum compression ([tex]KE_{final[/tex]) is:
[tex]KE_{\text{final}} = 0.5 \cdot (m_A + m_B) \cdot (v_{\text{common}})^2 = 0.5 \cdot 8.00 \, \text{kg} \cdot (0.50 \, \text{m/s})^2 = 1.00 \, \text{J} \\[/tex]The maximum energy stored in the spring bumpers ([tex]E_{spring[/tex]) is the difference between [tex]KE_{initial[/tex] and [tex]KE_{final[/tex]:
[tex]E_{\text{spring}} = KE_{\text{initial}} - KE_{\text{final}} = 4.00 \, \text{J} - 1.00 \, \text{J} = 3.00 \, \text{J} \\[/tex](b) Velocity of Each Block After Collision
After they have moved apart, assuming an elastic collision where both kinetic energy and momentum are conserved, the final velocities ([tex]v_A_{final[/tex] and [tex]v_B_{final[/tex]) can be found using the respective equations:
For block A:
[tex]v_{A_{\text{final}}} = \frac{(m_A - m_B)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(2 \cdot m_B)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{A_{\text{final}}} = \frac{(2.00 \, \text{kg} - 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(2 \cdot 6.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = -1.00 \, \text{m/s} \\[/tex]For block B:
[tex]v_{B_{\text{final}}} = \frac{(2 \cdot m_A)}{(m_A + m_B)} \cdot v_{A_{\text{initial}}} + \frac{(m_B - m_A)}{(m_A + m_B)} \cdot v_{B_{\text{initial}}} \\[/tex][tex]v_{B_{\text{final}}} = \frac{(2 \cdot 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 2.00 \, \text{m/s} + \frac{(6.00 \, \text{kg} - 2.00 \, \text{kg})}{(2.00 \, \text{kg} + 6.00 \, \text{kg})} \cdot 0 \, \text{m/s} = 1.00 \, \text{m/s} \\[/tex]Therefore, the final velocities are:
[tex]v_{A_{\text{final}}} = -1.00 \, \text{m/s} \quad \text{and} \quad v_{B_{\text{final}}} = 1.00 \, \text{m/s}[/tex]a wave is 8 meters long and has a frequency of 3 Hz. Find speed
Answer:
The speed is 24 [tex]\frac{meter}{s}[/tex]
Explanation:
A wave is a disturbance that propagates through a certain medium or in a vacuum, with transport of energy but without transport of matter.
The wavelength is the minimum distance between two successive points of the wave that are in the same state of vibration. It is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The speed of propagation is the speed with which the wave propagates in the middle, that is, the magnitude that measures the speed at which the wave disturbance propagates along its displacement. Relate wavelength (λ) and frequency (f) inversely proportionally using the following equation:
v = f * λ.
In this case, λ= 8 meter and f= 3 Hz
Then:
v= 3 Hz* 8 meter
So:
v= 24 [tex]\frac{meter}{s}[/tex]
The speed is 24 [tex]\frac{meter}{s}[/tex]
The speed of the wave is 24 m/s.
To find the speed of a wave, you can use the relationship between speed, frequency, and wavelength.
The formula is:
Speed (v) = Wavelength (λ) × Frequency (f)Given in this problem:
Wavelength (λ) = 8 metersFrequency (f) = 3 HzWe substitute these values into the formula:
Speed (v) = 8 meters × 3 HzThus, the speed of the wave is:
v = 24 m/sA very thin 19.0 cm copper bar is aligned horizontally along the east-west direction. If it moves horizontally from south to north at velocity = 11.0 m/s in a vertically upward magnetic field and B = 1.18 T , what potential difference is induced across its ends ? which end (east or west) is at a higher potential ? a) East b) West
Answer:
2.47 V,East
Explanation:
We are given that
l=19 cm=[tex]19\times 10^{-2} m[/tex]
[tex] 1 cm=10^{-2} m[/tex]
[tex]v=11 m/s[/tex]
B=1.18 T
We have to find the potential difference induced across its ends.
[tex]E=Bvl[/tex]
Using the formula
[tex]E=1.18\times 11\times 19\times 10^{-2}[/tex]
[tex]E=2.47 V[/tex]
Hence, the potential difference induces across its ends=2.47 V
The positive charge will move towards east direction and the negative charge will move towards west direction because the direction of force will be east.Therefore, the potential at east end will be high.
The induced potential difference in the copper bar is 2.47 V, with the east end being at a higher potential.
To determine the potential difference induced across the ends of a copper bar moving through a magnetic field, we use the formula:
V = B * l * v
where:
B is the magnetic field strength (1.18 T).l is the length of the bar (0.19 m).v is the velocity (11.0 m/s).Substituting the given values:
V = 1.18 T * 0.19 m * 11.0 m/s = 2.47 V
The potential difference across the ends of the bar is 2.47 V.
To determine which end is at a higher potential, we apply the right-hand rule. Pointing the thumb of your right hand in the direction of the velocity (north), and your fingers in the direction of the magnetic field (upwards), the palm points towards the force acting on positive charges (from west to east).
Therefore, the east end is at a higher potential.
The correct answer is: a) East
A 70.0-cm length of a cylindrical silver wire with a radius of 0.175 mm is extended horizontally between two leads. The potential at the left end of the wire is 3.15 V, and the potential at the right end is zero. The resistivity of silver is 1.586 ✕ 10−8 Ω · m.
a. What are the magnitude and direction of the electric field in the wire?
b. What is the resistance of the wire?
c. What are the magnitude and direction of the current in the wire?
d. What is the current density in the wire?
Answer:
(a) the magnitude and direction of the electric field in the wire is 4.5 N/C towards the left end of the wire.
(b) the resistance of the wire is 0.1154 Ω
(c) the magnitude and direction of the current in the wire is 27.3 A towards the left end of the wire
(d) the current density in the wire is 4.053 x 10⁸ A/m³
Explanation:
Given;
Length of cylinder = 70cm = 0.7m
radius of cylinder = 1.75 x 10⁻⁴ m
potential V = 3.15 V
resistivity = 1.586 ✕ 10⁻⁸ Ω · m
Part (a) the magnitude and direction of the electric field in the wire
V = E x d
E = V/d
E = 3.15/0.7 = 4.5 N/C towards the left end of the wire.
Part (b) the resistance of the wire
[tex]R = \frac{\rho L}{A} \\\\R =\frac{\rho L}{\pi r^2} =R = \frac{1.586X10^{-8} X 0.7}{\pi (1.75X10^{-4})^2} = 0.1154 ohms[/tex]
R = 0.1154 Ω
Part (c) the magnitude and direction of the current in the wire
V = IR
I =V/R
I = 3.15/0.1154
I = 27.3 A towards the left end of the wire
Part (d) the current density in the wire
current density,J = current /volume
volume = πr²h = π x (1.75 × 10⁻⁴)² x 0.7 = 6.7357 x 10⁻⁸ m³
[tex]J = \frac{27.3}{6.7357 X10^{-8}} = 4.053 X10^8 \frac{A}{m^3}[/tex]
J = 4.053 x 10⁸ A/m³
A liquid in a test tube has a curved surface such that the edges touching the glass are higher than the surface at the center. This must mean that the cohesive forces are less than the adhesive forces. 1. False 2. True
Answer:
True
Explanation:
Cohesive forces are forces that exist between the molecules of a substance of the same material while adhesive forces are forces that exists between the molecules of the substances of different materials.
This is responsible for the nature of menisci formed by different liquids when they are filled into containers or glass tubes.
If the cohesive forces of the liquid molecules are stronger than the adhesive forces between the liquid molecules and the glass material, a convex meniscus will be formed. This means that the edges of the liquid touching the glasses are lower than the surface at the centre. The meniscus formed by mercury in a glass tube is an example of this.
However, If the cohesive forces of the liquid molecules are weaker than the adhesive forces between the liquid molecules and the glass material, a concave meniscus will be formed. This means that the edges of the liquid touching the glasses are higher than the surface at the centre. The meniscus formed by water in a glass tube is an example of this.
Please view the attached diagram:
Please note: I got the diagram online, it was not drawn by me. I Just needed to quickly get something to illustrate my explanations. Thanks.
ball of mass 0.4 kg is attached to the end of a light stringand whirled in a vertical circle of radius R = 2.9 m abouta fixed point. Find the magnitude of the tension when themass is at the top if its speed at the top is 8.5 m/s.(
Answer:
[tex]6.046N[/tex]
Explanation:
The net force exerted on the mass is the sum of tension force and the external force of gravity.
[tex]F_n_e_t=F_g+F_t[/tex]
[tex]F_t[/tex] is the tension force.[tex]F_g=9.8N/kg[/tex] is the force of gravity.
[tex]F_n_e_t=ma_c=mv^2/r\\[/tex]
where [tex]r[/tex] is the rope's radius from the fixed point.
From the net force equation above:
[tex]F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N[/tex]
Hence the tension force is 6.046N
While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock and the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?
Answer:
μk = (Vf - Vc)/(T×g)
Explanation:
Given
Vi = initial velocity of the car
Vf = final velocity of the car
T = Time of application of brakes
g = acceleration due to gravity (known constant)
Let the mass of the car be Mc
Assuming only kinetic frictional force acts on the car as the driver applies the brakes,
The n from Newtown's second law of motion.
Fk = Mc×a
Fk = μk×Mc×g
a = (Vf - Vc)/T
Equating both preceding equation.
μk×Mc×g = Mc × (Vf - Vc)/T
Mc cancels out.
μk = (Vf - Vc)/(T×g)
Final answer:
To find the coefficient of kinetic friction of the tires against the road, you need the initial and final speeds of the car, the time it took to decelerate, and the acceleration due to gravity. Without specific values for these variables, the problem cannot be solved directly, but the formula given provides a method to calculate the coefficient if such values are known.
Explanation:
The question asks for the coefficient of kinetic friction between the tires and the road when a car, initially traveling at speed vi, decelerates to a final speed vf over time T seconds due to the driver slamming on the brakes upon seeing a rabbit.
The formula to calculate the coefficient of kinetic friction (μ_k) can be derived from Newton's second law of motion and the equation of motion that relates initial velocity, final velocity, acceleration, and time. The formula for the coefficient of kinetic friction is μ_k = (vi - vf) / (g × T), where g is the acceleration due to gravity (9.81 m/s2).
To solve this problem, you need the initial and final velocities of the car (vi and vf), the deceleration time (T), and knowledge that the acceleration due to gravity (g) is approximately 9.81 m/s2. However, actual calculations cannot be performed without specific values for vi, vf, and T.
This equation illustrates that the coefficient of kinetic friction is directly related to the deceleration rate of the car on the road surface.
"It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a body give away something it does not have in the first place
Answer:
Because heat is a path function or the energy in transit.
Explanation:
It is not correct to say that a body contains a certain amount of heat because the heat is a path function and not a property of the system. It is the energy in transit which can be encountered only when it crosses the system boundary.Heat is the energy in transit of a matter which flows by the virtue of temperature difference. The heat energy in a body is stored in the form of kinetic energy of the molecules which gets converted into heat that we know as the responsible factor for the rise in temperature usually.Heat is not a substance a body possesses; it's energy transfer. Describing heat as something a body "contains" is inaccurate. Heat is energy in transit during temperature differences.
The statement that a body contains a certain amount of heat is inaccurate in the context of thermodynamics. Heat is not a substance that a body can possess like a tangible quantity. Instead, heat is a form of energy transfer between systems due to a temperature difference. When we say a body transfers heat, it implies an exchange of thermal energy between the body and its surroundings.
A body does not "contain" heat in the way it contains mass or volume. Rather, it possesses internal energy, and heat is the energy in transit. This distinction is crucial in understanding the principles of thermodynamics. When a body transfers heat, it signifies a change in its internal energy, which could result from molecular motion or other energy interactions.
Therefore, the expression "giving away something it does not have" is a misconception. The body has internal energy, and during a heat transfer, this internal energy changes, affecting the body's temperature. It is crucial to frame discussions about heat in terms of energy transfer rather than possession, aligning with the principles of thermodynamics.
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The question probable may be:
It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a body give away something it does not have in the first place?
If you shine a laser on two slits with a separation of 0.215 mm, and the diffraction pattern shines on a background 4.90 m away, what is the wavelength of the laser light if the fringes are separated by 1.60 cm
Answer:
[tex]\lambda = 702\ nm[/tex]
Explanation:
Given,
slit width, d = 0.215 mm
slit separation, D = 4.90 m
Wavelength of the laser light = ?
Fringe width.[tex]\beta = 1.60\ cm[/tex]
Using formula
[tex]\beta =\dfrac{\lambda D}{d}[/tex]
[tex]\lambda=\dfrac{\beta d}{D}[/tex]
[tex]\lambda=\dfrac{0.016\times 0.215\times 10^{-3}}{4.90}[/tex]
[tex]\lambda = 7.02\times 10^{-9}\ m[/tex]
[tex]\lambda = 702\ nm[/tex]
Wavelength of the laser light = [tex]\lambda = 702\ nm[/tex]
The width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature across the oven show that there are cold spots at each edge of the oven and at three spots in between. The wavelength of the microwaves is 12 cm . How wide is the oven?
Answer:
Explanation:
In standing wave pattern we find region of nodes where vibration is minimum or cold spots . The distance between any two consecutive node is half the wave length . There are 5 cold spot or node in between which is equal to 4 half wave length .
width of oven = 4 x half wave length
= 4 x (12 / 2 )
= 24 cm
. A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the RC time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?
Answer:
initial current I₀ = 0.0123 A
RC time constant τ = 0.00075 sec
current after one time constant I = 0.00452 A
voltage on the capacitor after one time constant V = 3.89 V
Explanation:
Given that,
Voltage = 6.16 V
Resistance = 500 Ω
Capacitance = 1.5 µF
(a) What is the initial current?
The initial current can be found using
I₀ = Voltage/Resistance
I₀ = 6.16/500
I₀ = 0.0123 A
(b) What is the RC time constant?
The time constant τ provides the information about how long it will take to charge the capacitor.
τ = R*C
τ = 500*1.5x10⁻⁶
τ = 0.00075 sec
(c) What is the current after one time constant?
I = I₀e^(-τ/RC)
I = 0.0123*e^(-1) (0.00075/0.00075 = 1)
I = 0.00452 A
(d) What is the voltage on the capacitor after one time constant?
V = V₀(1 - e^(-τ/RC))
Where V₀ is the initial voltage 6.16 V
V = 6.16(1 - e^(-1))
V = 6.16*0.63212
V = 3.89 V
That means the capacitor will charge up to 3.89 V in one time constant
Answer:
Explanation:
Given an RC circuit to analyze
R=500Ω
C=1.50-μF uncharged
Emf(V)=6.16V
Series connection
a. Initial current, since the capacitor is initially uncharged then, the voltage appears at the resistor
Using ohms law
V=iR
Then, i=V/R
i=6.16/500
i=0.01232 Amps
i=12.32 mA.
b. The time constant is given as
τ=RC
τ=500×1.5×10^-6
τ=0.00075second
τ=0.75 ms
c. What is current after 1 time constant
Current in a series RC circuit is given as
Time after I time constant is
t=1 ×τ
t= τ
i=V/R exp(-t/RC)
Where RC= τ
i=V/Rexp(-t/ τ)
i=6.16/500exp(-1), since t= τ
i=0.004532A
I=4.532mA
d. Voltage after one time constant
Voltage of a series RC circuit(charging) is given as
Again, t= τ
V=Vo(1 - exp(-t/ τ)),
V=6.16(1-exp(-1))
V=6.16(1-0.3679)
V=6.16×0.632
V=3.89Volts
V=3.89V
V=
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.55.
To find the force necessary to start the crate moving, we use the formula fs(max) = μsN, where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. Plugging in the given values, the force necessary to start the crate moving is 169.36 N.
Explanation:The force necessary to start the crate moving is equal to the maximum static friction force. To calculate this, we use the formula:
fs(max) = μsN
where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the crate, which can be calculated by N = mg, where m is the mass of the crate and g is the acceleration due to gravity. So, the force necessary to start the crate moving is:
fs(max) = μsN = μsmg
Plugging in the given values:
fs(max) = (0.55)(32 kg)(9.8 m/s²) = 169.36 N
You are given a sample of several compounds to separate by paper chromatography. You draw a pencil line exactly 1.0 cm from the bottom of the paper, and place a spot of sample on it. You dry the sample, then develop it in a solvent. When the chromatogram is taken out of the solvent, the paper is wet up to 8.8 cm from the bottom of the sheet. The compound you are interested in shows up as a spot 7.5 cm from the bottom of the paper. Calculate the following: How far did the compound move?
Answer:
The compound moves 6.5 cm in total.
Explanation:
Before solving this problem, let's first write down all lengths we know of from the question:
Starting point of sample = 1.0 cm from bottom of paper
Paper wet up to = 8.8 cm from bottom of paper
Ending point of the sample = 7.5 cm from bottom of paper
With these lengths stated, we can easily calculate the length which the compound moved through:
Length compound moved = Ending point - Starting point
Length compound moved = 7.5 - 1.0
Length compound moved = 6.5 cm
Thus, we can see that the compound moved 6.5 cm between the time the paper was put into, and taken out of the solvent.
A 5.93 5.93 kg ball is attached to the top of a vertical pole with a 2.03 2.03 m length of massless string. The ball is struck, causing it to revolve around the pole at a speed of 4.25 4.25 m/s in a horizontal circle with the string remaining taut. Calculate the angle, between 0° and 90°, that the string makes with the pole. Take g = 9.81 g=9.81 m/s2.
Final answer:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.Therefore angle is approximately 45.6°.
Explanation:
To find the angle that the string makes with the pole, we can use the concept of centripetal force. The tension in the string provides the centripetal force that keeps the ball in circular motion.
At the top of the circle, the tension in the string is equal to the weight of the ball, which is given by T = mg. We can use this equation to find the angle the string makes with the pole.
tan(theta) = T/ (m*v²/ R), where m is the mass of the ball, v is the speed of the ball, and R is the length of the string. Plugging in the values given, we get tan(theta) = (5.93 kg * 4.25 m/s²) / (5.93 kg * 9.81 m/s²). Solving for theta, we find that the angle is approximately 45.6°.
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3, 550 J/kg·K, 48 W/m·K), which is initially at a uniform temperature of 170°C and is to be heated to a minimum temperature of 550°C. Heating is effected in a gas-fired furnace, where products of combustion at [infinity] 800°C maintain a convection coefficient of 250 W/m2·K on both surfaces of the plate. How long should the plate be left in the furnace?
Answer:
The answer to the question is;
The plate be left in the furnace for 905.69 seconds.
Explanation:
To solve the question, we have to check the Bi number as follows
Bi = [tex]\frac{hL}{k} = \frac{250\frac{W}{m^{2} K} *0.05 m}{48\frac{W}{mK} } = 0.2604[/tex]
As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.
We perform nonlumped analysis
The relation for heat transfer given by
Y = [tex]\frac{T_f-T_{inf}}{T_i- T_{inf}}[/tex]
=[tex]\frac{550-800}{170- 800}[/tex] = 0.3968 = C₁ exp (ζ₁² F₀)
where
C₁ and ζ₁ are coefficients of a series solution
We therefore look for the values of C₁ and ζ₁ from Bi tables to be
ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and
C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and
This gives the relation
0.3968 = 1.03956 exp (ζ₁² F₀)
or ζ₁² [tex](\frac{\alpha t}{L^2})[/tex]
where
α = Thermal diffusivity of solid = k/(ρ·c[tex]_p[/tex]) = [tex]\frac{48}{7830*550}[/tex] = 1.1146×10⁻⁵
c[tex]_p[/tex] = Specific heat capacity of solid at constant pressure = 550 J/kg·K
ρ = Density of the solid = 7830 kg/m³
=㏑[tex](\frac{0.3968 }{1.03956 })[/tex] = -0.9631 from where we have
t = [tex]\frac{0.9631 *0.05^{2} }{0.4884^2*1.11*10^{-5}}[/tex] = 905 seconds.
Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is elliptical?
Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.
This led to Kepler's law of orbital motion.
First Law: Planetary orbits are elliptical with the sun at a focus.
Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.
Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.
It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.
You have a device that takes temperature measurements and runs off of solar power. How often it is programmed to take a measurement will affect how much power it uses--more frequent measurements, more power. You have one installed at the equator and one installed in the Antarctic. During which season of the year can you set each device to take more frequent measurements
Answer:
In the Equator:
As far as the temperature is concerned equator is more or less the same throughout the year, however, there are some fluctuations also, I will set this device in March and September because, in this month, the Sun is exactly over the equator and I would be able to get the more results in this month.
In Antarctic:
As far as the climatic conditions of Antarctic are concerned, it is all the same while it fluctuates in December because the Sun is very much close to Antarctic that's why I will choose this month.
Explanation:
The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2 × 10-12 e-940/T, where T is the Kelvin temperature. Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C. Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3.
Answer:
Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -50 °C = 223 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]
Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH
Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3
Knowing
Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x[tex]10^{-11} e^{-255/T}[/tex] and 2x[tex]10^{-12} e^{-940/T}[/tex]
T = -80 °C = 193 K
The reaction rate will be given by [Cl] [O3] 3x[tex]10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3][/tex]
Than, the reaction rate of OH with O3 is
Rate = [OH] [O3] 2x[tex]10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3][/tex]
Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]
Than, considering the concentration of Cl increases by a factor of 100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex], the result will be that the reaction with OH will be 535 + (100 to about 4 × [tex]10^{5}[/tex] molecules [tex]cm^{-3}[/tex]) times faster than the reaction with Cl
Explanation:
An ice cube and a rubber ball are both placed at one end of a warm cookie sheet, and the sheet is then tipped up. The ice cube slides down with virtually no friction, and the ball rolls down without slipping. The ball and the ice cube have the same inertia. Which one reaches the bottom first?
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
What is the friction force?It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. its unit is Newton (N).
Mathematically, it is defined as the product of the coefficient of friction and normal reaction.
The ice ball reaches the bottom first, when the sheet is then tipped up. In the given condition, the ball and the ice cube have the same inertia.
Because the ice cube slides down with virtually no friction. Without friction, an object can easily slide with the more speed.
The ball is spherical in nature and a spherical surface is in more contact with the surface during the inclined motion.
Hence, the ice ball reaches the bottom first, when the sheet is then tipped up.
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The speed of sound in air is 345 m/s. A tuning fork vibrates above the open end of a sound resonance tube. If sound waves have wavelengths 58-cm in the tube, what is the frequency (in Hertz) of the tuning fork
Answer:
594.8 Hz
Explanation:
Parameters given:
Speed of sound, v = 345 m/s
Wavelength = 58 cm = 0.58 m
Speed of a wave is given as:
Speed = wavelength * frequency
Therefore:
Frequency = Speed/Wavelength
Frequency = 345/0.58
Frequency = 594.8 Hz
The frequency of the tuning fork is calculated using the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is wavelength. Substituting their given values, we get a frequency of approximately 595 Hz.
Explanation:
The frequency of sound can be calculated with the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is the wavelength.
From your question, we know that the speed of sound (Vw) in air is 345 m/s and the wavelength (a) is 58 cm (or 0.58 m when converted to meters to match the speed of sound's units).
Frequency (f) can be calculated by rearranging to f = Vw / a. Substituting the values: f = 345 m/s / 0.58 m = 594.83 Hz. So, the tuning fork's frequency is approximately 595 Hz.
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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. The same cart again traveling at 0.3 m/s collides with a different stationary object. This time the cart is at rest after the collision. In which collision is the impulse on the cart greater?A. The impulses are the same.B. The second collision.C. The first collision.D. Cannot be determined without knowing the mass of the cart.E. Cannot be determined without knowing the rebound speed of the first collision.
Answer: Impulse is greater in the first case. So, option C is the correct option.
Explanation:
Case 1: Cart is travelling at 0.3 m/s and collide with an stationary object and after collision, cart rebound in opposite direction and another object remains in static condition.
Applying the conservation of linear momentum:
[tex]m_{1} \times u_{1} + m_{2} \times u_{2} = m_{1} \times v_{1} + m_{2} \times v_{2}[/tex]
[tex]m_{1} \times 0.3 + m_{2} \times 0 = m_{1} \times v_{1} + m_{2} \times 0[/tex]
Hence velocity of cart will rebound with the same velocity i.e. 0.3 m/s
Impulse is defined as the change in momentum
Impulse on the cart = [tex]m_{1} \times v_{1} - m_{1} \times u_{1}[/tex] = [tex]m_{1} \times ((-3) - (3)) = m_{1} \times (-6)[/tex] Kg m/s.
Case 2: Initially cart is travelling at 0.3 m/s and after collision it comes to rest.
So, change in momentum or Impulse = [tex]m_{1} \times (0 - 3)[/tex] = [tex]-3 \times m_{1}[/tex] Kg m/s.
Impulse is greater in the first case. So, option C is the correct option.
Answer:
The second impulse is greater then the first impulse.
(B) is correct option.
Explanation:
Given that,
In first case,
Initial speed of cart = 0.3 m/s
Final speed of cart = -0.3 m/s
in second case,
Initial speed of cart = 0.3 m/s
Final speed of cart = 0 m/s
In first case,
We need to calculate the impulse
Using formula of impulse
[tex]I=\Delta p[/tex]
[tex]I=\Delta (mv)[/tex]
[tex]I=mv-mu[/tex]
Put the value into the formula
[tex]I=m(-0.3-0.3)[/tex]
[tex]I= -0.6m\ kg m/s[/tex]
In second case,
We need to calculate the new impulse
Using formula of impulse
[tex]I'=\Delta (mv')[/tex]
[tex]I'=mv'-mu'[/tex]
Put the value into the formula
[tex]I'=m(0-0.3)[/tex]
[tex]I'= -0.3m\ kg m/s[/tex]
So, I'> I
Hence, The second impulse is greater then the first impulse.
The vertical displacement of an ocean wave is described by the function, y = A sin(ωt - kx). k is called the wave number (k = 2π/λ) and has a value of k = 18 rad/m. The remaining values are A = 9.5 m and ω = 14.5 rad/s.
a) Using y = A sin(B), input an expression for B where the wave would be traveling in the -x-direction. sig.gif?tid=7M79-31-9F-4E-8624-20536
b) What is the wave's velocity in m/s?
c) What is the wave's amplitude in m?
The expression for B for a wave traveling in the -x direction is B = ωt - kx + π. The wave velocity is 0.805 m/s and the wave amplitude is 9.5 m.
Explanation:To determine the expression for B where the wave would be traveling in the -x direction, we need to consider that the general equation for the wave function is y = A sin(B). In this case, the wave is traveling in the -x direction, which means the phase of the wave is shifted by π (180 degrees). So the expression for B would be B = ωt - kx + π.
The wave's velocity can be calculated using the formula v = ω/k. Substituting the given values, the wave's velocity is v = 14.5 rad/s / 18 rad/m = 0.805 m/s.
The wave's amplitude is given directly as A = 9.5 m.
A U-shaped tube open to the air at both ends contains water. A quantity of oil of unknown density is slowly poured into the right arm of the tube until the vertical height of the oil column is 20cm. The top of the oil is 8cm higher than the top of the water. Find the density of the oil.
Answer:
[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil
Explanation:
Given:
height of oil column, [tex]h_o=20\ cm[/tex]oil column height that is more than the water column height in the other arm, [tex]\delta h=8\ cm[/tex]Now from the given it is clear that the height of water column is:
[tex]h_w=h_o-\delta h[/tex]
[tex]h_w=20-8[/tex]
[tex]h_w=12\ cm[/tex]
Now according to the pressure balance condition of fluid columns:
Pressure due to water column = Pressure due to oil column
[tex]P_w=P_o[/tex]
[tex]\rho_w.g.h_w=\rho_o.g.h_o[/tex]
[tex]1000\times 9.8\times 0.12=\rho_o\times 9.8\times 0.2[/tex]
[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil
Answer:
Explanation:
Let the density of oil is d'.
height of water, h = 20 - 8 = 12 cm
height of oil, h' = 20 cm
density of water, d = 1000 kg/m³
Pressure is balanced
h' x d' x g = h x d x g
0.20 x d' x g = 0.12 x 1000 x g
0.2 d' = 120
d' = 600 kg/m³
In the lab, you did not include friction in your calculations for the acceleration. Explain why it was not necessary. What would your equation for acceleration look like if you were to include friction
Final answer:
Friction was not included in the lab calculations because it opposes motion between surfaces, resulting in a smaller acceleration. The equation for acceleration without friction is a = g sinθ, while the equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction.
Explanation:
The reason friction was not included in the calculations for acceleration in the lab is because friction always opposes motion between surfaces, resulting in a smaller acceleration when it is present. In the absence of friction, all objects slide down a frictionless incline with the same acceleration, regardless of mass. The equation for acceleration without friction is a = g sinθ, where g is the acceleration due to gravity and θ is the angle of the incline.
If friction were to be included in the calculations, the equation for acceleration would be different. It would depend on the coefficient of friction (μ) and the normal force (N) of the object. The equation for acceleration with friction is a = (g sinθ - μk cosθ) / (1 + μk sinθ), where μk is the coefficient of kinetic friction. This equation takes into account the opposing force of friction and provides a more accurate representation of the object's acceleration.
You have a 78.7 mF capacitor initially charged to a potential difference of 11.5 V. You discharge the capacitor through a 3.03 Ω resistor. What is the time constant?
Answer:
[tex]\tau=0.23\;second[/tex]
Explanation:
Given,
[tex]C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\[/tex]
Time constant
[tex]\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second[/tex]
Determine the effective spring constant of the suspension system of a car. Consider a load of 4 passengers, each with a mass of 70 kg. The car suspension system consists of four identical springs, each with a spring constant k. When all four people get into the car, the tires are depressed by about Δx = 2.0 cm.
Answer:
The value of spring constant for each spring of the suspension system of a car K = 343.35 [tex]\frac{N}{cm}[/tex]
Explanation:
Total force on the springs = weight of the four passengers
⇒ F = 4 × 70 × 9.81
⇒ F = 2746.8 N
In the suspension system of the car the four springs are connected in parallel. So Equivalent spring constant is given by,
⇒ [tex]K_{eq}[/tex] = 4 K -------- ( 1 )
Depression in the spring Δx = 2 cm
Now the force on the spring is given by
F = [tex]K_{eq}[/tex] × Δx
⇒ [tex]K_{eq}[/tex] = [tex]\frac{2746.8}{2}[/tex]
⇒ [tex]K_{eq}[/tex] = 1373.4 [tex]\frac{N}{cm}[/tex]
Now the spring constant for each spring = [tex]\frac{K_{eq}}{4}[/tex]
⇒ K = [tex]\frac{1373.4}{4}[/tex]
⇒ K = 343.35 [tex]\frac{N}{cm}[/tex]
This is the value of spring constant for each spring of the suspension system of a car.
The effective spring constant of the suspension system of a car. is mathematically given as
K = 343.35N/cm
What is the effective spring constant of the suspension system of a car.?
Question Parameter(s):
Consider a load of 4 passengers, each with a mass of 70 kg.
the tires are depressed by about dx = 2.0 cm.
Generally, the equation for the Force is mathematically given as
F = Keq × dx
Therefore
Keq= 2746.8/2
Keq= 1373.4N/cm
In conclusion, spring constant for each spring
K = 1373.4/4
K = 343.35N/cm
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A small ranger vehicle has a soft, ragtop roof. When the car is at rest, the roof is flat. When the car is cruising at highway speeds with its windows rolled up, does the roof bow upwards, remain flat, or bow downwards
Answer:
roof bow upwards
Explanation:
The top of the roof of the small ranger vehicle will bow upwards. This is as a result of gas pressure on the soft ragtop roof.
As air begins to fill the vehicle, pressure resonates in all direction proportionally. The pressure of the air will be greater than that which the roof can withstand and this forces the roof sky up. It is a common scene when we see roof of ragtop vehicles bowing upwards into the sky.Which statement below is NOT true about electric field lines?
A) They can start on + charges
B) They are perpendicular to the electric field at every point
C) They get closer together where the field is stronger
D) Field lines cannot cross
E) They can end on - charges
Assuming the passive sign convention and an operating frequency of 314 rad/s, calculate the phasor voltage V which appears across each of the following when driven by the phasor current I = 1020° mA: (a) a 2 a resistor; (b) a 1 F capacitor; (c) a 1 Hinductor; (d) a 2 resistor in series with a 1 F capacitor; (e) a 2. resistor in series with a 1 H inductor. (f) Calculate the instantaneous value of each voltage determined in part (e) at t = 0.
Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment
You give a book a brief push to make it slide up a rough inclined surface. It comes to a stop and slides back down to the starting point.
Does it take the same amount of time to go up as it does to come down?
Answer:No, it will take a longer time to slide up than to slide down.
Explanation: An inclined plane is a plane that slides at an angle to the ground,this angle can be 45 degrees,60degrees etc. WHEN AN OBJECT IS SLIDING UP A ROUGH INCLINED SURFACE, IT WILL HAVE TO OVERCOME THE FRICTION OR RESISTANCE OF THE ROUGH SURFACE AND THE OPPOSITION OF THE FORCE IF GRAVITY, WHICH WILL CAUSE IT TO TAKE A LONGER TIME TO GO UP WHILE WHEN SLIDING DOWN IT WILL ONLY HAVE TO OVERCOME THE FRICTION OF THE ROUGH SURFACE AS THE FORCE OF GRAVITY ACTS TO PULL IT DOWN THE SLOPE.