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3 years ago

8

a single 1,300 jk cargo car is rolling along a train track at 2.0 m/s when 400 kg of coal is dropped vertically into it. what is

it’s velocity right afterward? assume a closed system.

1 answer:

tensa zangetsu [6.8K]3 years ago

4 0

Answer:

1.53 m/s

Explanation:

Given:

Mass of the car (M) = 1300 kg

Mass of the coal (m) = 400 kg

Initial velocity of the car (U) = 2 m/s

Initial velocity of the coal (u) = 0 m/s (Since it is dropped)

When the coal is dropped into the car, then they move with same final velocity.

Let the final velocity be 'v' m/s.

For a closed system, the law of conservation of momentum holds true.

So, initial momentum is equal to final momentum of the car-coal system.

Initial momentum of the car = $MU=1300\times 2=2600\ Ns$

Initial momentum of the coal = $mu=0\ Ns$

Total initial momentum is the sum of the above two momentums.

So, total initial momentum = 2600 + 0 = 2600 Ns

Now, final momentum is given as the product of combined mass and final velocity. So,

Final momentum of the system = $(M+m)v=(1300+400)v=1700v$

Now, from law of conservation of momentum,

Initial momentum = Final momentum

$2600=1700v\\\\v=\frac{2600}{1700}\\\\v=1.53\ m/s$

Therefore, the final velocity of either of the two masses is same is equal to 1.53 m/s.

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8 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

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Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

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The average velocity is total displacement divided by time:

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$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

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The angles in the triangle are 91 degrees, 53 degrees and 36 degrees respectively.

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From the cosine rule we know that;

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c = 0.78 m

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(0.78)^2 = (0.47)^2 + (0.62)^2 - 2(0.47 * 0.62)cosC

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0.01 = - 0.58 cosC

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Learn more about triangle:brainly.com/question/2773823

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6 0

1 year ago