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2 years ago

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A 17-kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 33 N. Starting from rest, the sle

d attains a speed of 1.6 m/s in 9.8 m. Find the coefficient of kinetic friction between the runners of the sled and the snow.

1 answer:

ycow [4]2 years ago

4 0

Answer:

$\mu=0.185$

Explanation:

From the question we are told that:

Mass $m=17kg$

Force $F=33N$

Velocity $v=1.6m/s$

Distance $d= 9.8m$

Generally the equation for Work done is mathematically given by

$W=\triangle K.E+\triangle P.E$

Where

$\triangle K.E=(F-F_f)*2$

$F_f=F+\frac{\triangle K.E}{d}$

$F_f=33+\frac{0.5*17*1.6^2}{9.8}$

$F_f=30.8N$

Since

$f = \mu*m*g$

$\mu= 30.8/(m*g)$

$\mu= 30.8/(17*9.81)$

$\mu=0.185$

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Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

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F cos θ = Ff

F cos 36 = 65 N

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<span>So the nurse must exert 80.34 N of force</span>

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3 years ago

Each of the following diagrams shows a spaceship somewhere along the way between Earth and the Moon (not to scale); the midpoint

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Answer:

$F_5 >F_4>F_1 >F_2>F_3$

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Explanation:

For this case the figure attached shows the illustration for the problem

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Th formula is given by:

$F = G \frac{m_{Earth} m_{Spaceship}}{r^2}$

Where G is a constant $G = 6.674 x10^{-11} m^2/ (ks s^2)$

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$m_{spaceship}$ represent the mass for the spaceship

$r$ represent the radius between the earth and the spaceship

For this reason when the distance between the Earth and the Spaceship increases the Force of gravity needs to decrease since are inversely proportional the force and the radius, and for the other case when the Earth and the spaceship are near then the radius decrease and the Force increase.

Based on this case we can create the following rank:

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3 years ago

What is Vatiable Velocity?

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3 years ago

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Answer:

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FBA = −FAB

However, this can be explained by Newton's second law. Let's say the truck has mass M and the car has mass m. If the magnitude of the force that both vehicles experience is F, then the magnitudes of their respective accelerations are:

atruck = F/M

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and combining these we get:

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So if the mass of the car is a lot less than the mass of the truck, then the acceleration of the truck is much smaller than the acceleration of the car, and if you were to watch the collision, the truck would pretty much seem like it's motion was unaffected, but the car's motion will change quite a bit.

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3 years ago

A 25kg box in released on a 27° incline and accelerates down the incline at 0.3 m/s2. Find the friction force impending its moti

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Answer:

a) μ = 0.475 , b) μ = 0.433

Explanation:

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the friction force has the expression

fr = μ N

y Axis

N - $W_{y}$ = 0

let's use trigonometry for the components the weight

sin 27 = Wₓ / W

Wₓ = W sin 27

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very = tan 27 - a / g sec 27

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