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Sveta_85 [38]
3 years ago
11

A 2.0-kg cart collides with a 1.0-kg cart that is initially at rest on a low-friction track. After the collision, the 1.0-kg car

t moves to the right at 0.50 m/s and the 2.0-kg cart moves to the right at 0.25 m/s . If the positive direction is to the right, what was the initial velocity of the 2.0-kg cart?
Physics
1 answer:
JulsSmile [24]3 years ago
4 0

Answer:

Initial velocity of first cart will be 0.5 m /sec

Explanation:

We have given mass of the cart m_1=2kg

Its collides with other cart of mass m_2=1kg

Let the initial velocity of cart 1 is v_{1i}

As the car 2 is initially at rest so v_{2i} = 0 m/sec

After the collision velocity of first cart is 0.25 m/sec

So v_{1f} = 0.25 m/sec and velocity of second cart is 0.50 m /sec

So v_{2f} = 0.5 m/sec

From conservation of momentum we know that

m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}

2\times v_{1i}+1\times 0=2\times 0.25+1\times 0.5

v_{1i} = 0.5 m /sec

So initial velocity of first cart will be 0.5 m /sec

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Oduvanchick [21]

Answer:

the average force exerted on the ball by the bat is 11,613.27 N

Explanation:

Given;

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time of action, t = 1.10 ms = 1.10 x 10⁻³ s

The average force exerted on the ball by the bat is calculate as;

F = ma = \frac{m(v-u)}{t} \\\\F = \frac{0.151(45.1-(-39.5))}{1.10\times 10^{-3}} \\\\F = \frac{0.151(45.1\ +\ 39.5)}{1.10\times 10^{-3}} \\\\F = 11,613.27 \ N

Therefore, the average force exerted on the ball by the bat is 11,613.27 N

7 0
2 years ago
The maximum energy a bone can absorb without breaking is surprisingly small. For a healthy human of mass 60 kg60 kg, experimenta
netineya [11]

Answer:

<em>the maximum height a man can jump from and land rigidly upright on both feet without breaking his legs is 0.34 m</em>

<em></em>

Explanation:

Mass of a healthy man = 60 kg

energy the bone can take without breaking = 200 J

If a healthy man jumps from a height 'h', he falls with an energy equal to the potential energy due to his initial height above the ground.

initial potential energy of the healthy man = mgh

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g = acceleration due to gravity = 9.81 m/s^2

h = the height above ground

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equating, we have

200 = 588.6h

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When people jump from a height, the sudden deceleration to zero can impact a big force on the leg bones, shattering them. If the time spent in decelerating to zero is increased, the overall force on the leg bones is reduced greatly.

<em>Bending the knees gradually on landing from a jump from a height, and then rolling increases the time spent decelerating, and reduces the impact force on the legs due to the landing</em>. If you observe carefully you will see that this is what professional stunts men and acrobats do when they jump from a height.

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Answer: question 1 is b I believe

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