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2 years ago

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A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and

the speed is 10 m/s. At this instant the cord is 3.40 m long. (Take the x direction to be horizontal and to the right, the y direction to be upward, and the z direction to be out of the page.)
At this instant, what is the magnitude of the rate of change of the child's momentum?

1 answer:

snow_tiger [21]2 years ago

7 0

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

$\dfrac{dp}{dt}=F$

$\dfrac{dP}{dt}=\dfrac{mv^2}{r}$

Put the value into the formula

$\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}$

$\dfrac{dP}{dt}=794.11\ N$

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

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A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i

sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, $v_{Ci}=25\ m/s$

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, $v_{Ti}=20\ m/s$

After the collision, velocity of the car, $v_{Cf}=18\ m/s$

Let $v$ is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

$initial\ momentum=final\ momentum$

$1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v$

$210000-21600=9000\ kg\times v$

$v=20.93\ m/s$

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

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1 year ago

The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan

PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire. If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

Magnetic field around a long current carrying wire is given by

$B=\frac{\mu _o I}{2\pi r}$

where B= magnetic field

$\mu _o=$ permeability of free space

I= current in the long wire and

r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u> proportional to the current and <u><em>inversely</em></u> proportional to the distance from the wire.

Now if I'=3I and r'=2r then magnetic field B' is given by

$B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B$

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

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2 years ago

Read 2 more answers

If you used 1000 J of energy to throw a ball, would it travel faster if you threw the ball (ignoring air resistance)

wolverine [178]

To solve this problem it is necessary to apply the kinematic equations of Energy for which the rotation of a circular body is described as

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

Where,

m = Mass of the Vall

v = Velocity

I = Moment of inertia abouts its centre of mass

$\omega =$ Angular speed

Basically the two sums of energies is the consideration of translational and rotational kinetic energy.

a. so that it was also rotating?

The ball is rotating means that it has some angular speed:

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

When there is a little angular energy (and not linear energy to travel faster), translational energy will be greater than the 1000J applied.

$1000J > \frac{1}{2}mv^2$

The ball will not go faster.

c. so that it wasn't rotating?

For the case where the angular velocity does not rotate it is zero therefore

$KE = \frac{1}{2}mv^2+\frac{1}{2}I\omega^2$

$1000J = \frac{1}{2}mv^2+\frac{1}{2}I(0)^2$

$1000J = \frac{1}{2}mv^2$

All energy is transoformed into translational energy so it is possible to go faster. This option is CORRECT.

b. It makes no difference.

Although the order presented is different, I left this last option because as we can see with the previous two parts if there is an affectation regarding angular movement, therefore it is not correct.

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1 year ago

The mass m1 enters from the left with velocity v0 and strikes a mass m2 > m1 which is initially at rest. The collision betwee

enot [183]

Answer:

1. False 2) greater than. 3) less than 4) less than

Explanation:

1)

As the collision is perfectly elastic, kinetic energy must be conserved.
The expression for the final velocity of the mass m₁, for a perfectly elastic collision, is as follows:

$v_{1f} = v_{10} *\frac{m_{1} -m_{2} }{m_{1} +m_{2}}$

As it can be seen, as m₁ ≠ m₂, v₁f ≠ 0.

2)

As total momentum must be conserved, we can see that as m₂ > m₁, from the equation above the final momentum of m₁ has an opposite sign to the initial one, so the momentum of m₂ must be greater than the initial momentum of m₁, to keep both sides of the equation balanced.

3)

The maximum energy stored in the in the spring is given by the following expression:

$U =\frac{1}{2} *k * A^{2}$

where A = maximum compression of the spring.

This energy is always the sum of the elastic potential energy and the kinetic energy of the mass (in absence of friction).
When the spring is in a relaxed state, the speed of the mass is maximum, so, its kinetic energy is maximum too.
Just prior to compress the spring, this kinetic energy is the kinetic energy of m₂, immediately after the collision.
As total kinetic energy must be conserved, the following condition must be met:

$KE_{10} = KE_{1f} + KE_{2f}$

So, it is clear that KE₂f < KE₁₀
Therefore, the maximum energy stored in the spring is less than the initial energy in m₁.

4)

As explained above, if total kinetic energy must be conserved:

$KE_{10} = KE_{1f} + KE_{2f}$

So as kinetic energy is always positive, KEf₂ < KE₁₀.

4 0

1 year ago

A muon formed high in the Earth's atmosphere is measured by an observer on the Earth's surface to travel at speed V - 0.983c for

Alex_Xolod [135]

Answer:

The moun lives 2.198*10^-6 s as measured by its own frame of reference

The Earth moved 648 m as measured by the moun's frame of reference

Explanation:

From the point of view of the observer on Earth the muon traveled 3.53 km at 0.983c

0.983 * 3*10^8 = 2.949*10^8 m/s

Δt = d/v = 3530 / 2.949*10^8 = 1.197*10^-5 s

The muon lived 1.197*10^-5 s from the point of view of the observer.

The equation for time dilation is:

$\Delta t' = \Delta t * \sqrt{1 - \frac{v^2}{c^2}}$

Then:

$\Delta t' = 1.197*10^-5 * \sqrt{1 - \frac{(0.983c)^2}{c^2}} = 2.198*10^-6 s$

From the point of view of the moun Earth moved at 0.983c (2.949*10^8 m/s) during a time of 2.198*10^-6, so it moved

d = v*t = 2.949*10^8 * 2.198*10^-6 = 648 m

7 0

1 year ago