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3 years ago

11

A sleigh weighing 2000 newtons is pulled my a horse a distance of 1.0 kilometer (or 1000 meters) in 45 minutes. what is the powe

r if the horse?

1 answer:

hoa [83]3 years ago

8 0

Work = Force* Distance
2000*1000=2000000

Power = Work/Time

2000000/45=<span>44444.44 Watts</span>

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a man exerts 700 newtons of force to move a piece of furniture 4 meters. if it takes him 2 seconds to move the furniture, how mu

dybincka [34]

   Work = (force) x (distance)

The work he did: Work = (700 N) x (4m) = 2,800 joules

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3 years ago

Suppose a particle moves along a straight line with velocity v(t)=t2e−2tv(t)=t2e−2t meters per second after t seconds. How many

dimulka [17.4K]

Explanation:

It is given that,

Velocity of the particle moving in straight line is :

$v(t)=t^2e^{-2t}\ m/s$

We need to find the distance (x) traveled by the particle during the first t seconds. It is given by :

$x=\int\limits {v.dt}$

$x=\int\limits {t^2e^{-2t}dt}$

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4 years ago

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

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3 years ago