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3 years ago

Rank the following in terms of increasing momentum:

Physics

2 answers:

lys-0071 [83]3 years ago

7 0

Answer:

C. a 1200 kg car going 15 m/s

D. a 15 kg meteor going at a speed of 1000 m/s

B. a 100 kg person running at 5 m/s

A. a 10,000 kg train car at rest

Explanation:

Momentum is the amount of kinetic energy that an object has, it depends on the amount of mass and the velocity at which the object is moving, so if you have an object at rest, it would have zero momentum, so the one with the most momentum would be the heaviest quickiest, and that one would be a 1200 kg car at 15 ms. then the 15 kg meteor at 1000 ms, then the 100 kg person at 5 m/s and last the train car at rest.

ycow [4]3 years ago

6 0

Answer:

Ranks based on momentum are

1. 1200 kg car going at 15 m/s.

2. 15 kg meteor going at a speed of 100 m/s.

3. 100 kg person running at 5 m/s.

4. 10000 kg train car at rest.

Explanation:

Product of a body’s mass and velocity gives the value of its momentum. It is given by the equation

$P=mv$

where is the mass of the object and is the velocity.

Calculating the values of momentum:

1. mass= 1200 kg, velocity= 15 m/s

$p=mv =1200 \times 15 = 18000 \ kg \ m/s$

2. mass= 15 kg, velocity= 1000 m/s

$p= 15 \times 1000 = 15000 \ kg \ m/s$

3. mass= 100 kg, velocity= 5 m/s

$p= 100 \times 5 = 500 kg \ m/s$

4. mass= 1000 kg, velocity= 0 m/s

$p= 10000 \times 0= 0 \ kg\ m/s$

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2 years ago

A worker at the top of a 588-m-tall television transmitting tower accidentally drops a heavy tool. If air resistance is negligib

liq [111]

The final velocity of the tool is 107.4 m/s

Explanation:

We can solve this problem by using the principle of conservation of energy.

In fact, if air resistance is negligible, the total mechanical energy of the tool is conserved during the fall, so we can write:

$K_i + U_i = K_f + U_f$

where

$K_i = 0$ is the kinetic energy of the tool at the top (zero since it is at rest)

$U_i = mgh$ is the gravitational potential energy of the tool at the top, where

m is the mass of the tool

g is the acceleration of gravity

h is the heigth of the tool

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the tool just before hitting the ground, where

v is the final speed of the tool

$U_f = 0$ is the gravitational potential energy of the tool at the bottom (zero since the height is zero)

Re-arranging the equation,

$mgh=\frac{1}{2}mv^2$

where we have

$g=9.8 m/s^2\\h=588 m$

And solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(588)}=107.4 m/s$

Learn more about kinetic energy and potential energy:

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3 years ago

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vladimir2022 [97]

Answer:

0.050V

Explanation:

To solve this problem it is necessary to use the concepts related to the potential between two objects that have a magnetic field, this concept is represented in the equation.

$\int dV = \int_{0}^{l/2} Bv (dl)$