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3 years ago

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A cart for hauling ore out of a gold mine has a mass of 435 kg, including its load. the cart runs along a straight stretch of tr

ack that is sloped 4.77Â° from the horizontal. a donkey, trudging along and to the side of the track, has the unenviable job of pulling the cart up the slope with a 438-n force for a distance of 193 m by means of a rope that is parallel to the ground and makes an angle of 13.3Â° with the track. the coefficient of friction for the cart\'s wheels on the track is 0.0167. use g = 9.81 m/s2. find the work that the donkey performs on the cart during this process.

1 answer:

marta [7]3 years ago

4 0

What are the answer choices

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120n

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An over-caffeinated student stands on a table which weighs 600 newtons. The student has a mass of 50 kg. What is the weight of t

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An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

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Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

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We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

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$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

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3 years ago

What is the force acting on a 10kg object that accelerates from 5 m/s to 20 m/s in 5s?

vitfil [10]

Answer:

Option C

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3 years ago

A 1.2 kg ball drops vertically onto a floor from a height of 32 m, and rebounds with an initial speed of 10 m/s.

iren [92.7K]

Explanation:

Given that,

Mass of the ball, m = 1.2 kg

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

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$J=m(v-u)$

$J=1.2\times (-25.04-10)$

J = -42.048 kg-m/s

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$J=\dfrac{F}{t}$

$F=J\times t$

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Hence, this is the required solution.

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3 years ago