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Colt1911 [192]
2 years ago
15

A box sits at the top of an incline that is 50.0 cm tall and that makes an angle of 25° with the ground. The box has a mass of

2.0 kg. It is given an initial speed of 3.0 m/s down the incline. If the coefficient of kinetic friction between the box and the incline is 0.10, how long does it take the box to reach the bottom of the incline?
Physics
1 answer:
DiKsa [7]2 years ago
7 0

As the box sits at the top of an incline that is 50.0 cm tall and makes an angle of 25° with the ground, The time it takes to reach the bottom of the incline is mathematically given as

t=0.148s

<h3>What is the time it takes the box to reach the bottom of the incline?</h3>

Generally, the equation for the acceleration  is mathematically given as

a=g(sin\theta-uscos\theta)

Therefore

a=(9.8sin-(0.15cos35))

a=4.417

Thereofore

v^2=v0^2+2al

v^2=V0^2+2ahsin\theta\\\\V^2=(2.0+2(4.417))*60*0.01sin35

v=2.6533

In conclusion, The time t is

t=v-v0/a

2.65-20/4.417

t=0.148s

Read more about  Arithmetic

brainly.com/question/22568180

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A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
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Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

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Now,

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Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

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On putting the values, we get

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Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

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On putting the values, we get

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hence,

The time taken by the boat will be:

= 26.08+35.29

= 61.37 \ s

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G to ml,density=3.291 g/ml
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