Explanation:
There is no any direct relationship between the frequency of a sound relative the intensity of the sound.
Frequency is a perception of pitch of the sound whereas intensity determines the loudness of the sound. Intensity is proportional to amplitude of the sound.
[tex]\beta= log_{10}\frac{I}{I_o}[/tex]
So, it would be hard to determine how much more intense is a 6000 Hz tone than a 80 Hz tone.
A 47 gram golf ball is driven from the tee with an initial speed of 52 m/sec and rises to a height of 24.6 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8 m below its highest pint
Answer:
a) 52.2 J
b) 48.77 m/s
Explanation:
a)47 g = 0.047 jg
The kinetic (and total mechanical energy) of the ball at the ground is
[tex]E = mv^2/2 = 63.544 J[/tex]
The potential energy of the ball at its highest point 24.6m is. Let g = 9.8m/s2
[tex]E_h = mgh = 0.047*9.8*24.6 = 11.33 J[/tex]
Since the potential energy at the highest height is less than the total mechanical energy on ground, the difference must be kinetic energy
[tex]E_k = E - E_h = 63.544 - 11.33 = 52.2J[/tex]
b) 8m below 24.6m is 16.6m. The potential energy at this point is
[tex]E_{p8} = mgh = 0.047*9.8*16.6 = 7.64 J[/tex]
And so the kinetic energy at this point is
[tex]E_{k8} = E - E_{p8} = 63.544 - 7.64 = 55.9 J[/tex]
So the speed is
[tex]mv^2/2 = 55.9[/tex]
[tex]v^2 = 2*55.9/0.047 = 2378.64[/tex]
[tex]v = \sqrt{2378.64} = 48.77 m/s[/tex]
The kinetic energy of the golf ball at its highest point is zero and the speed of the ball when it is 8 m below its highest point can be determined using the principle of conservation of mechanical energy.
Explanation:To determine the kinetic energy of the golf ball at its highest point, we can use the principle of conservation of energy. Since the ball is at its highest point, its gravitational potential energy is maximum, and its kinetic energy is minimum. Therefore, the kinetic energy of the golf ball at its highest point is zero.
To find the speed of the ball when it is 8 m below its highest point, we can use the principle of conservation of mechanical energy. We can equate the initial mechanical energy of the ball to its final mechanical energy. The initial mechanical energy is the sum of the initial kinetic energy and the initial potential energy. The final mechanical energy is the sum of the final kinetic energy and the final potential energy. By solving this equation, we can find the speed of the ball when it is 8 m below its highest point.
Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the driver of car A accelerates his car with a uniform forward acceleration of 2.40 m/s2. How long after car A begins to accelerate does it take car A to overtake car B?A) 12.6 s B) 5.50 s C) 19.0 s D) 316 s E) Car A never overtakes car B.
Answer:
(c) 19.0s
Explanation:
Given Data
Car A speed v=22.0 m/s
Car B speed v=29.0 m/s
Car A distance S=300 m behind Car B
Car A acceleration a=2.40 m/s²
To find
Time required For Car A to take over Car B
Solution
We can represent Car A Coordinate by using equation of simple motion
[tex]X_{A} =vt+1/2at^{2}\\X_{A} =22t+(1/2)(2.40)t^{2}[/tex]
And Coordinates of car B equals
[tex]X_{B}=300+29t\\[/tex]
Car A is overtake car B when:
[tex]X_{A}=X_{B}\\ 22t+(1/2)(2.4)t^{2}=300+29t\\1.2t^{2}-7t-300=0\\ time=19.0s[/tex]
Option (C) 19.0s is correct one
By setting the distance traveled by car A equal to the distance traveled by car B plus the initial separation and solving the resulting quadratic equation, it is determined that car A will overtake car B in 20 seconds.
Explanation:To find out how long it takes for car A to overtake car B, we must calculate the respective positions of the cars as a function of time and find when they match. Car A is accelerating from an initial speed, while car B is moving at a constant speed.
The position of car A as a function of time can be found with the equation:
sA = vA0t + ½at2
Where:
The position of car B as a function of time (since car B is not accelerating) is simply:
sB = vBt
Where:
Car A will overtake car B when sA equals sB plus the initial 300 m separation. We can set up the equation and solve for t.
22t + ½(2.40)t2 = 29t + 300
Rearranging to solve for t gives a quadratic equation:
1.2t2 - 7t - 300 = 0
Using the quadratic formula or factoring, we solve for t. It yields t = 20 s (after discarding the negative solution which is not physically meaningful).
Therefore, it takes car A 20 seconds to overtake car B, which results in Option C being the correct answer.
1. A woman driving at the 45 mi/hour speed limit on the entrance ramp to the highway accelerates at a constant rate and reaches the highway speed limit of 65 mi/hour in 6.00 s. What distance does the car travel during that acceleration? (Make the simplifying assumption that she is traveling in a straight line and be careful with your units)
Answer:
s = 147.54 m
Explanation:
given,
initial velocity,u = 45 mi/h
1 mph = 0.44704 m/s
45 mph = 45 x 0.44704 = 20.12 m/s
final velocity, v = 65 mi/h
v = 65 x 0.44704 = 29.06 m/s
time, t = 6 s
acceleration, [tex]a = \dfrac{v-u}{t}[/tex]
[tex]a = \dfrac{29.06-20.12}{6}[/tex]
a = 1.49 m/s²
distance travel by the car
using equation of motion
v² = u² + 2 a s
29.06² = 20.12² + 2 x 1.49 x s
2.98 s = 439.6692
s = 147.54 m
distance traveled by the car is equal to 147.54 m
Consider three force vectors Fi with magni- tude 53 N and direction 116º, F2 with mag- nitude 57 N and direction 217°, and F3 with magnitude 71 N and direction 20°. All direc- tion angles 0 are measured from the positive x axis: counter-clockwise for 0 > 0 and clock- wise for 0 < 0. What is the magnitude of the resultant vec- tor || F ||, where F = Fi + F2 + 3 ? Answer in units of N. 004 (part 2 of 2) 10.0 points What is the direction of É as an angle between the limits of -180° and +180° from the positive x axis with counterclockwise as the positive angular direction? Answer in units of 005 10.0 points Consider the instantaneous velocity of a body. This velocity is always in the direction of 1. the least resistance at that instant. 2. the net force at that instant. 3. the motion at that instant.
To find the magnitude and direction of the resultant vector, use the Pythagorean theorem and inverse tangent function respectively.
Explanation:To find the magnitude of the resultant vector, we can use the Pythagorean theorem. The resultant vector is the sum of the three force vectors: F1, F2, and F3. We can find the x and y components of each vector using trigonometry, and then add the components to find the x and y components of the resultant vector. Finally, we can use the Pythagorean theorem to find the magnitude of the resultant vector.
To find the direction of the resultant vector, we can use the inverse tangent function to find the angle between the positive x-axis and the resultant vector. Since the problem specifies that angles are measured counterclockwise from the positive x-axis, we need to make sure the angle is within the range of -180° to +180°. If the angle is greater than 180°, we subtract 360° to get the equivalent angle within the specified range.
The magnitude of the resultant vector is 95.2 N and the direction is -95.5°.
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A 52 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. The acceleration of gravity is 9.8 m/s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.
Answer:
6.1 m
Explanation:
m = Mass of person = 52 kg
h = Altitude
v = Velocity
Kinetic energy of the person on the ground
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}52\times 11^2\\ =3146\ J[/tex]
Kinetic energy of the person at the top
[tex]\dfrac{1}{2}52\times 1.2^2\\ =37.44\ J[/tex]
At the top the potential energy is given by
[tex]mgh=52\times 9.8h[/tex]
Balancing the energy of the system
[tex]3146=37.44+52\times 9.8h\\\Rightarrow h=\dfrac{3146-37.44}{52\times 9.8}\\\Rightarrow h=6.1\ m[/tex]
Her altitude is 6.1 m
Stefan's Law says:a. the energy radiated by a blackbody is proportional to T³. c. the hotter a star's surface, the bluer it looks to us. E = mc². d. that if the Sun's temperature were doubled, it would give off 16X more energy. d. that doubling the star's temperature would also double its peak wavelength.
Answer:
d. that if the Sun's temperature were doubled, it would give off 16X more energy
Explanation:
Stefan's Law ;
According to this law ,the energy emitted by a body which have temperature T(in K) is proportional to the forth power of the temperature .This is apllicable for the black bodies(These bodies absorb all the incident radiation or we can say that these are perfect absorber bodies).
[tex]Energy\ \alpha\ T^4\\E \alpha\ T^4\\[/tex]
If we double the temperature then the energy will become 16 times of the initial energy.
That is why the option d is correct.
An oceanic depth-sounding vessel surveys the ocean bottom with ultrasonic waves that travel at 1530 m/s in seawater. The time delay of the echo to the ocean floor and back is 6 s. ?
Answer:
d = 4590 m
Explanation:
given,
Speed of ultrasonic wave, v = 1530 m/s
time of the echo, t = 6 s
Let d be the depth of the ocean
now,
total distance travel by the ultrasonic wave = 2 d
we know,
distance = speed x t
2 d = 1530 x 6
d = 1530 x 3
d = 4590 m
Hence, the depth of the ocean floor is equal to d = 4590 m
A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s later. You may ignore air resistance. Part A Part complete If the height of the building is 21.0 m m , what must the initial speed be of the first ball if both are to hit the ground at the same time? v v = 9.53 m/s m/s SubmitPrevious Answers Correct Part B Part complete Consider the same situation, but now let the initial speed v 0 v0 of the first ball be given and treat the height h h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v 0 v0v_0 = 8.70 m/s m/s . h h = 6.51 m m SubmitPrevious Answers CorrectPart C Part complete If v 0 v0 is greater than some value v max vmax , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v max vmax . v max vmax = 11.7 m/s m/s SubmitPrevious Answers Correct Part D Part complete If v 0 v0 is less than some value v min vmin , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v min vmin . v min vmin = 5.83 m/s m/s SubmitPrevious Answers Correct Provide Feedback Next
A) Initial velocity of ball 1: 9.53 m/s upward
B) Height of the building: 6.48 m
C) Maximum velocity: 11.7 m/s
D) Minimum velocity: 5.83 m/s
Explanation:
A)
The y-position of the 1st ball at time t is given by the equation for free fall motion:
[tex]y_1 = h + v_0 t - \frac{1}{2}gt^2[/tex] (1)
where
h = 21.0 m is the initial height of the ball, the height of the building
[tex]v_0[/tex] is the initial velocity of the ball, upward
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by
[tex]y_2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
where
h = 21.0 m is the initial height of the ball, the height of the building
t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then [tex]y_2=h[/tex], so the ball is still on the roof
The 2nd ball reaches the ground when [tex]y_2=0[/tex], so:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0[/tex]
Which has two solutions:
t = -0.88 s (negative, we discard it)
t = 3.26 s (this is our solution)
The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and [tex]y_1=0[/tex], so we find the initial velocity:
[tex]0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s[/tex]
B)
In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:
[tex]v_0 = 8.70 m/s[/tex]
When the two balls reach the ground at the same time, there position is the same, so we can write:
[tex]y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
Solving the equation, we find:
[tex]v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s[/tex]
This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m[/tex]
C)
If [tex]v_0[/tex] is greater than some value [tex]v_{max}[/tex], then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}[/tex]
becomes negative: in that case, the time becomes negative, so no solution is possible.
The denominator becomes negative when
[tex]1.19g-v_0 < 0[/tex]
Therefore when
[tex]v_0>1.19g=(1.19)(9.8)=11.7 m/s[/tex]
So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.
D)
There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19[/tex]
Solving the equation for [tex]v_0[/tex], we find:
[tex]0.5g(1.19)^2>1.19(1.19g-v_0)\\6.94>13.88-1.19v_0\\1.19v_0>6.94[/tex]
Which gives
[tex]v_0>5.83 m/s[/tex]
Therefore, the minimum speed of ball 1 at the beginning must be 5.83 m/s.
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Kirchhoff's Rules When applying Kirchhoff's rules, one of the essential steps is to mark each resistor with plus and minus signs to label how the potential changes from one end of the resistor to the other. The circuit in the drawing contains four resistors, each marked with the associated plus and minus signs. However, one resistor is marked incorrectly. Which one is it? a.R1 b.R2 c.R3 c.R4
Answer:
d. R4
Explanation:
Generally, the flow of current is always from the positive sign to the negative sign. In the resistors R1, R2, and R3, the direction of flow of current is from the positive sign to the negative sign. However, in the resistor R4, the direction of the flow of current is different from the conventional method. Therefore, the resistor R4 is marked wrongly.
Resistance R₄ is wrongly connected in the circuit.
Current always flow in the opposite direction of flow of electrons.
Since, Resistor acts as load in the electrical circuit. So, it will always consumed power.
Thus, In resistor current always flow from higher potential to lower potential i.e. from positive terminal to negative terminals.
In given circuit, current flowing in resistances R₁, R₂ and R₃ , from positive to negative. But in resistance R₄ current is flowing from negative to positive.
Therefore, Resistance R₄ is wrongly connected in the circuit.
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If a bucket with water is hung to a spring balance and then a weight with the help of a thread is fully immersed inside the bucket such that it does not touch sides of the bucket, will the spring balance reading change or remain same?
Answer:
Explanation:
The weight fully immersed will displaced water of equal volume to itself. The weight of this water displaced = volume of water displaced × density of water × acceleration due to gravity. This increase in weight will lead the spring balance reading to change; increase because there is an increase in mass.
The difference between the full wave rectifier with the center tap transformer, and the full wave bridge rectifier is_______________. Question 9 options: in full wave bridge rectifier two didoes conduct during each half of the cycle whereas one diode conducts during each half cycle in the center tapped full wave rectifier. transformer configuration is simpler in the full wave bridge rectifier. that the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Answer:
That the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Explanation:
See Image
Final answer:
In the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. The peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
Explanation:
The difference between the full wave rectifier with the center tap transformer and the full wave bridge rectifier is that in the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. Additionally, the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
if a material has a half-life of 24 hours, how long do you have to wait until the amount of radioisotope is 1/4 its original amount?
Answer:
48 hours
Explanation:
Using the formula,
R/R' = 2ᵃ/ᵇ..................... Equation 1
Where R = Original amount, R' = Radioactive remain, a = Total time, b = half life.
Given: b = 24 hours,
Let: R = X, then R' = X/4.
Substitute into equation 1
X/(X/4) = 2ᵃ/²⁴
4 = 2ᵃ/²⁴
2² = 2ᵃ/²⁴
Equating the base and solving for a
2 = a/24
a = 24×2
a = 48 hours.
Hence the time = 48 hours
The half-life of a material is the time it takes for half of the atoms in a sample to decay. If a material has a half-life of 24 hours, then after 48 hours (i.e., two half-lives), the amount of the material will be 1/4 of its original amount.
Explanation:The half-life concept applied in this question falls under the subject of Physics, specifically nuclear physics. The half-life of a radioactive material is the time it takes for half of the atoms in a sample to decay. If a material has a half-life of 24 hours, then after 24 hours, half of the original material would remain.
Given this, if we want the amount of the material to be only 1/4 of its original amount, we simply wait for another half-life. Remember, each half-life reduces the original amount by half. So after the first 24 hours (the first half-life), half of the material would have decayed. If you wait another 24 hours (another half-life), half of what remained after the first half-life would decay again, leaving you with 1/4 of the original amount.
So, you would have to wait 48 hours for the amount of radioisotope to be 1/4 of its original amount assuming the half-life is 24 hours.
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___ twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
Answer:
shielded
Explanation:
shielded twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:_______
a- doubled
b- unchanged
c- quadrupled
d- quartered
e- halved
Answer:e
Explanation:
Given
Area of parallel plates is A
distance between plates is d
Potential difference between Plates is V
Capacitance is given by
[tex]C=\frac{\epsilon _0A}{d}[/tex]
If separation is doubled then capacitance become half
[tex]C'=\frac{\epsilon _0A}{2d}[/tex]
[tex]C'=\frac{C}{2}[/tex]
Electrical energy stored in the capacitor is given by
[tex]E=\frac{1}{2}CV^2[/tex]
When distance is doubled
[tex]E'=\frac{1}{2}\times \frac{C}{2}\times V^2[/tex]
[tex]E'=\frac{E}{2}[/tex]
Therefore Energy is halved
If the separation between the plates is doubled, the electrical energy stored in the capacitor will be: d. The electrical energy stored in the capacitor will be quartered.
To understand why the energy stored in the capacitor is quartered when the separation between the plates is doubled, let's consider the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor.
The capacitance C of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]
The energy U stored in a capacitor is given by:
[tex]\[ U = \frac{1}{2} C V^2 \][/tex]
[tex]\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \frac{\varepsilon_0 A}{d} = \frac{1}{2} C \][/tex]
[tex]\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{1}{2} C\right) V^2 = \frac{1}{4} C V^2 = \frac{1}{4} U \][/tex]
So, when the separation between the plates is doubled, the capacitance is halved, and since the energy is directly proportional to the capacitance, the energy stored in the capacitor is quartered.
What scenarios best describes how the hawaiian islands formed in the pacific ocean?
Answer:
Magma generated from a hot spot burned through the overlying plate to create volcanoes.
Explanation:
The Earth’s outer crust is made up of a series of tectonic plates that move over the surface of the planet. In areas where the plates come together, volcanoes will form in most cases. Volcanoes could also form in the middle of a plate, where magma rises upward until it erupts on the seafloor which is called a hot spot.
Hawaiian Islands were formed by such a hot spot occurring in the middle of the Pacific Plate. While the hot spot itself is fixed, the plate is moving. As the plate moved over the hot spot, the string of islands that make up the Hawaiian Island were formed.
Brownian motion is due to:
a. The random movement of pollen granules suspended in water.
b. The random fluctuation of the energy content of the environment. Thermal noise.
c. The random fluctuation of the energy content of the environment and thermal noise.
Answer:
option B
Explanation:
Brownian motion is the random movement of the microscopic size particle suspended in a liquid or gas.
Brownian motion is the result of the collision of the fast-moving particles.
This phenomenon was described by Robert Brown in the year 1927 i.e. it is named Brownian motion.
Brownian motion is due to the random fluctuation of the energy in the environment which leads to the zig-zag movement of the Particle.
Hence, the correct answer is option B.
Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
Answer:
Gravitational potential energy
Explanation:
Gravitational potential energy is the type of energy an object has due to its position in a gravitational field. Water behind a dam possesses gravitational potential energy due to it being at a higher level than the water on the other side of the dam. When the water falls the gravitational potential energy is converted to kinetic energy, leading to the turning of the turbines to generate electricity.
In a very large closed tank, the absolute pressure of the air above the water is 6.46 x 105 Pa. The water leaves the bottom of the tank through a nozzle that is directed straight upward. The opening of the nozzle is 2.86 m below the surface of the water. (a) Find the speed at which the water leaves the nozzle. (b) Ignoring air resistance and viscous effects, determine the height (relative to the submerged end of nozzle) to which the water rises.
Answer:
a) 35.94 ms⁻²
b) 65.85 m
Explanation:
Take down the data:
ρ = 1000kg/m3
a) First, we need to establish the total pressure of the water in the tank. Note the that the tanks is closed. It means that the total pressure, Ptot, at the bottom of the tank is the sum of the pressure of the water plus the air trapped between the tank rook and water. In other words:
Ptot = Pgas + Pwater
However, the air is the one influencing the water to move, so elimininating Pwater the equation becomes:
Ptot = Pgas
= 6.46 × 10⁵ Pa
The change in pressure is given by the continuity equation:
ΔP = 1/2ρv²
where v is the velocity of the water as it exits the tank.
Calculating:
6.46 × 10⁵ =1/2 ×1000×v²
solving for v, we get v = 35.94 ms⁻²
b) The Bernoulli's equation will be applicable here.
The water is coming out with the same pressure, therefore, the equation will be:
ΔP = ρgh
6.46 × 10⁵ = 1000 x 9.81 x h
h = 65.85 meters
Sabe-se que um alqueire paulista equivale a 24200 metros quadrados. Uma chácara retangular tem um alqueire e mede 100m de frente. Quanto ela mede de fundo?
Answer:
b = 242 m
Explanation:
A = 24200 m²
a = 100 m
b = ?
A seguinte fórmula é aplicada
A = a*b
⇒ b = A / a
⇒ b = (24200 m²) / (100 m)
⇒ b = 242 m
An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 50, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Answer:
The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
Explanation:
Given that,
Mass of object = 2.5 kg
Momentum [tex]p= 3i+6j+7k[/tex]
Force [tex]F=50i+50j+100k[/tex]
Time [tex]t=5\times10^{-3}\ s[/tex]
We need to calculate the momentum of the object at the end
Using formula of impulse
[tex]J=\Delta p[/tex]
[tex]J=m\Delta v[/tex]...(I)
[tex]J=F\times\Delta t[/tex]....(II)
From equation (I) and (II)
[tex]F\times\Delta t=m\Delta v[/tex]
[tex]F\times \Delta t=m(v_{f}-v_{i})[/tex]
[tex]mv_{f}=F\times \Delta t+mv_{i}[/tex]
Put the value into the formula
[tex]p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k[/tex]
[tex]p_{f}=0.25i+0.25j+0.5k+3i+6j+7k[/tex]
[tex]p_{f}=(3.25i+6.25j+7.5k)\ kg m/s[/tex]
Hence, The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
To determine the object's momentum after the force is applied, calculate the change in momentum from the applied force over the time interval and add it to the initial momentum. The final momentum of the object is <3.25, 6.25, 7.5> kg·m/s.
Explanation:The student asked: What is the momentum of the object at the end of this time interval? To calculate the final momentum, we must use the formula p = p_0 + F ∙ Δt, where p_0 is the initial momentum, F is the force applied, and Δt is the time interval.
With the given values:
Initial momentum: p_0 = <3, 6, 7> kg·m/sForce: F = <50, 50, 100> NTime: Δt = 5 × 10^-3 sCalculating the change in momentum due to the force:
Change in momentum: Δp = F ∙ Δt = <50, 50, 100> N ∙ 5 × 10^-3 s = <0.25, 0.25, 0.5> kg·m/sAdding the initial momentum to the change in momentum gives the final momentum:
Final momentum: p = <3, 6, 7> + <0.25, 0.25, 0.5> = <3.25, 6.25, 7.5> kg·m/sTherefore, the object's momentum at the end of the time interval is <3.25, 6.25, 7.5> kg·m/s.
The water molecule has a dipole with the negative portion
A) localized on one of the hydrogens
B) localized between the hydrogen atoms
C) pointing toward the oxygen atom
D) pointing from the oxygen through the hydrogen atoms
In a water molecule, the dipole's negative portion is located on the oxygen atom and points towards the hydrogen atoms due to the difference in electronegativity between these elements.
Explanation:The water molecule, H2O, is a polar molecule, meaning it has a net dipole due to the presence of polar bonds, which result from a significant difference in electronegativities of the atoms involved. In a water molecule, the oxygen atom is more electronegative than hydrogen and therefore pulls the shared electrons toward itself. This creates a charge separation where the oxygen side of the molecule becomes partially negative, and the hydrogen side becomes partially positive. Looking at the provided options, the most accurate statement would be that the negative portion of the dipole is D) pointing from the oxygen through the hydrogen atoms
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A space vehicle is traveling at 5320 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 98 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation ?
Answer:
5398.4km/h
Explanation:
IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.
Let mass of module be m
then
mass of motor = 4m (four times the mass of rocket module)
total mass = m + 4m = 5m
combined velocity = V = 5320kph
Let
absolute (relative to earth)motor velocity after disengagement = v
then
rocket module velocity (relative to earth) after disengagement = v+98 (relative velocity = 98)
momentum conservation equation
combined momentum = module momentum + motor momentum
(m+4m)V = m(v+98) + 4m*v
5mV = 98m+mv + 4mv
5V = 98+v + 4v (m cancels out)
5V - 98 = 5v
((5*5320)-98)/5 = v
v = 5300.4 km/h
velocity of rocket module relative to earth = v +98
= 5300.4 + 98
= 5398.4km/h
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.
PART A)
Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinitePART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
The Electric Field Inside a Conductor is always zero, and the charge density inside a conductor is always zero. The charge density on the surface of the conductor can be expressed as eta = E / epsilon_0, where E is the magnitude of the electric field just outside the surface of the conductor and epsilon_0 is the permittivity of free space.
The electric field inside a conductor placed in an external electrostatic field is always zero (option d). This is because when an external electric field is applied to a conductor, the charges inside the conductor rearrange themselves in such a way that the net electric field inside the conductor becomes zero.
The charge density inside the conductor is also always zero (option a). In equilibrium, the charges in a conductor reside on its surface, creating an electric field within the conductor that is zero.
If the electric field just outside the surface of the conductor has a magnitude E and is directed toward the surface, the charge density (eta) on the surface of the conductor can be expressed as eta = E / (epsilon_0), where epsilon_0 is the permittivity of free space.
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(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, ν, and a versus t for the interval from one start-up to the next.
Answer:
The correct answer is 32.9 m/s
Explanation:
To solve this, we list out the known and the unknown variables as follows
Maximum allowable acceleration = 1.34 m/s²
Distance between sttions = 806 m
Therefore from the equation of motion
v = ut + 0.5·×at²
Where v = final velocity
u = initial velocity
S = distance covered
t = time
a = acceleration
Also v² = u² + 2·a·S
where u is the initial velocity, which we can take as u = 0, then
v² = 2·1.34·S = 2.68S m²/s² then
Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z, so u² = 2.68Z
since u² = 2.68S from the previous calculation, then for v = 0
2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m
and v² = 2.68S m²/s² = 1080.04 m²/s²
v = 32.9 m/s
The maximum speed a subway train can attain between stations is 32.9 m/s
The maximum speed a subway train can attain between stations is approximately 51.73 m/s. The travel time between stations is about 15.6 seconds. The maximum average speed of the train, from one start-up to the next, is approximately 27.62 m/s.
Explanation:To determine the maximum speed a subway train can attain between stations, we need to use the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Plugging in the values, we have v² = 0 + 2(1.34 m/s²)(806 m). Solving for v, we find that the maximum speed the subway train can attain between stations is approximately 51.73 m/s.
The travel time between stations can be calculated using the equation t = s/v, where t is the time, s is the displacement, and v is the velocity. Plugging in the values, we have t = 806 m / 51.73 m/s ≈ 15.6 seconds.
The maximum average speed of the train, from one start-up to the next, is given by the equation v_avg = (2s)/(t + t_stop), where v_avg is the average speed, s is the displacement, t is the travel time, and t_stop is the time the train stops at each station. Plugging in the values, we have v_avg = (2 * 806 m) / (15.6 s + 20 s) ≈ 27.62 m/s.
To graph x, ν, and a versus t for the interval from one start-up to the next, we would need specific values for x and a over time. However, we can consider a simplified case where a is constant, resulting in a linear graph of a versus t. The graph of v versus t would have a constant positive slope equal to the acceleration, and the graph of x versus t would be a curve with a positive concave upwards shape.
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During a lunar mission, it is necessary to increase the speed of a spacecraft by 2.76 m/s when it is moving at 400 m/s relative to the Moon. The speed of the exhaust products from the rocket engine is 1100 m/s relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?
Explanation:
Formula to calculate initial mass of given spacecraft is as follows.
[tex]v_{f} - v_{i} = v_{rel} \times ln(\frac{M_{i}}{M_{f}})[/tex]
The given data is as follows.
[tex]v_{f} - v_{i}[/tex] = 2.76 m/s
[tex]v_{rel}[/tex] = 1100 m/s
[tex]\frac{M_{f}}{M_{i}} = e^\frac{-dv}{v_{rel}}[/tex]
So, [tex]\frac{M_{i} - M_{f}}{M_{i}} = 1-e^-(\frac{2.76}{1100})[/tex]
= [tex]2.51 \times 10^{-3}[/tex]
Thus, we can conclude that [tex]2.51 \times 10^{-3}[/tex] fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase.
An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.
Answer:
(a) Total area is 14.5 roods
(b) Total area is 14674.522 square meters
Explanation:
Area occupied by land = 3 acres
1 acre = 40 perches by 4 perches = 160 square perches
3 acres = 3×160 = 480 square perches
Area occupied by livestock = 25 perches by 4 perches = 100 square perches
Total area = 480 + 100 = 580 square perches
1 rood = 4 perches by 1 perch = 4 square perches
580 square perches = 580/4 = 14.5 roods
(b) Total area = 580 square perches
1 perch = 16.5ft = 16.5/3.2808 = 5.03 meters
580 square perches × (5.03 meters/1 perch)^2 = 580 ×25.3009 square meters = 14674.522 square meters
The total area owned by the landowner is 580 perches, which is equivalent to 14.5 roods. When converted to the modern unit of measurement, this totals to approximately 17516 square meters.
Explanation:To solve this problem, we first need to convert each area to the same unit. In this case, we can convert all areas to perches. According to the old manuscript, the landowner had 3 acres of plowed land. Given that 1 acre is equivalent to 160 perches (40 perches by 4 perches), the plowed land was 480 perches (3 x 160).
The livestock area is 25 perches by 4 perches, which totals to 100 perches. Therefore, the total area owned by the landowner is 580 perches (480 perches + 100 perches).
(a) To convert this to roods, we divide by 40 (since 1 rood is 40 perches by 1 perch) which equals to 14.5 roods
(b) To convert perches to square meters, we need to know that 1 perch is 16.5 ft. and 1 square foot is approximately 0.0929 square meters. So, 1 perch is 325.125 sq ft (16.5 ft * 16.5 ft). Therefore, 1 perch is about 30.2 square meters (325.125 ft * 0.0929), and 580 perches equate to approximately 17516 square meters.
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Marcus used a toaster oven in the morning.He notices that when he plug it in and turn it on the coils inside begin to glow red what transformation are taking place
Answer:Conversion of electric energy to Heat energy
Explanation:Energy is a quantitative energy measured in JOULES or KILOJOULES which must be transferred to a material for a job to be done. It has also been described as the capacity to do work.
In electric toasters the ELECTRIC ENERGY FROM THE SOURCE IS TRANSFERRED INTO THE TOASTER TO BE CONVERTED TO HEAT ENERGY NEEDED TO TOAST FOODS. Other electrical appliances which converts electric energy to Heat energy includes ELECTRIC BOILERS, ELECTRIC COOKERS etc.
A certain spring is found not to obey Hooke’s law; it exerts a restoring force Fx(x)=−αx−βx2 if it is stretched or compressed, where α=60.0N/m and β=18.0N/m2. The mass of the spring is negligible. (a) Calculate the potential-energy function U(x) for this spring. Let U = 0 when x = 0. (b) An object with mass 0.900 kg on a frictionless, horizontal surface is attached to this spring, pulled a distance 1.00 m to the right (the +x-direction) to stretch the spring, and released. What is the speed of the object when it is 0.50 m to the right of the x = 0 equilibrium position?
Answer:U(x) = 30x^2 +6x^3
V^2=8.28m/s
Explanation:The law of conservation of energy is given by K1+U1= K2+U2 ...eq 1
Kinetic energy K.E= 1/2 mv^2
Restoring force function F(x)= -60x - 18x^2
But F(x)= -dU/dx
dU(x)=-F(x)dx
Integrating U(x)= -integral F(x)dx + U(0)
Substituting, we get
U(x) = - integral(-60x-18x^2)dx+U(0)
U(x)= 30x^2+6x^3+U(0)
U=0 at x=0
Therefore U(x)= 30x^2+6x^3
b) Given : x1=1.00m,x2= 0.50m ,V1=0, V2=?
Substituting into eq (a)
U1= 30(1.00)^2+6(1.00)^3=36J
Using x2=0.5 into eq(a)
U2=30(0.50)^2+6(0.50)^3=8.25J
Object at rest K1=0
0+36=K2+8.25
K2=27.75J
Given; m =0.900kg, V2=?
27.75=1/2×0.900×V2^2
V2= SQRT(2×27.75)/0.81
V2= 8.28m/s
The relationship between force, potential energy and energy conservation allows to find the results for the questions about the spring are:
a) The potential energy is: U = 30 x² + 6 x³
b) The velocity is: v = 7.85 m / s
Given parameters.
Restorative force f = - α x - β x² Constants values α = 60.0 N / m and β = 18.0 N / m² Body mass m = 0.900 kg Displacement initial x₁ = 1,0 m and final x₂ = 0,5 mTo find
a) Potential energy.
b) Speed.
a) Force and potential energy are related by the expression.
[tex]F = - \frac{dU}{dx}[/tex]
Where F is the force and U the potential energy.
dU = - F dx
∫ dU = - ∫∫ (-α x - β x²) dx
U- U₀ = [tex]\alpha \frac{x^2}{2} + \beta \frac{x^3}{3}[/tex]alpha / 2 x² + beta / 3 x³
Let's substitute the constants values and indicate that U₀=0 when x=0.
U = [tex]30 x^2 + 6x^3[/tex]
b) They ask the speed of the block between two points, as they indicate that there is no friction we can use the theorem of conservation of mechanical energy, which states that energy is conserved at all points.
Starting point.
Em₀ = U (1)
Final point.
[tex]Em_f[/tex] = K + U (0.5)
Energy is conserved.
Em₀ = Em_f
U (1) = K + U (0.5)
Where the kinetic energy is
K = ½ m v²
Let's substitute.
v² = [tex]\frac{2}{m} [ U(1) - U(0.5)][/tex]
Let's calculate.
v² = [tex]\frac{2}{0.900}[/tex] [ (30 1² + 6 1³) - (30 0.5² + 6 0.5³) ]
v = [tex]\sqrt{\frac{2 \ 27.75 }{ 0.900} }[/tex]
v = 7.85 m / s
In conclusion using the relationship between force, energy potential and the conservation of energy we can find the results for the questions about the spring are:
a) The potential energy is: U = 30 x² + 6 x³
b) The velocity is: v = 7.85 m / s
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In a population of Mendel's garden peas, the frequency of dominant yellow-flowered plants is 50%. The population is in Hardy-Weinberg equilibrium. What is the frequency of the homozygous recessive genotype in the population? 0.71 0.25 0.5 The frequency cannot be determined from the data provided.
The frequency of the homozygous recessive genotype in the population is approximately 0.71.
Explanation:The frequency of the homozygous recessive genotype in the population can be calculated using the Hardy-Weinberg equation. In this case, the frequency of the dominant yellow-flowered plants is given as 50% or 0.5. To calculate the frequency of the homozygous recessive genotype (qq), we need to find the value of q. The equation for the Hardy-Weinberg equilibrium is p² + 2pq + q² = 1.
Given that the frequency of the dominant allele (p) is 0.5, we can substitute the value of p into the equation and solve for q:
0.5² + 2(0.5)(q) + q² = 1
Simplifying the equation gives:
0.25 + q + q² = 1
Combining like terms:
q² + q - 0.75 = 0
Using the quadratic formula to solve for q, we find that q ≈ 0.71 or -1.21. Since the frequency of a trait cannot be negative, we can disregard the negative value. Therefore, the frequency of the homozygous recessive genotype in the population is approximately 0.71.
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Calculate the potential energy of a 300 gram volleyball that is 15 meters in the air. (g = 9.8 m/s2)
explain why if you can, thank you :)
Answer:
The book gravitational potential energy is 44.1 Joules fear
Explanation:
Given:
Mass of volleyball, m = 300g = 300 ÷ 1000 = 0.3 kg
height, h = 15 m
To Find:
Potential Energy = ?
Solution:
Gravitational Potential Energy :
Gravitational potential energy is energy an object possesses because of its position in a gravitational field.
Formula is given by
[tex]\textrm{Gravitational Potential Energy}= m\times g\times h[/tex]
Where,
m = mass
g = acceleration due to gravity = 9.8 m/s²
h = height
Substituting the values we get
[tex]\textrm{Gravitational Potential Energy}= 0.3\times 9.8\times 15\\\\\textrm{Gravitational Potential Energy}=44.1 Joules[/tex]
The book gravitational potential energy is 44.1 Joules.
The potential energy of a 300 gram volleyball that is 15 meters in the air is 44.1 Joules.
To calculate the gravitational potential energy (PEg), you would use the formula PEg = mgh,
where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.
Since the mass of the volleyball is given as 300 grams, you must first convert this mass into kilograms, resulting in 0.3 kg. Then, use the given values with g = 9.8 m/s² to find the potential energy:
PEg = (0.3 kg) × (9.8 m/s2) × (15 m)
PEg = 4.41 kg×m²/s²
PEg = 44.1 Joules