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2 years ago

9

Two dogs are pulling a disk, but the disk is not moving. The disk does not move because the forces are (select one: unbalanced,

balanced, in motion, at rest.)

2 answers:

tiny-mole [99]2 years ago

8 0

Answer:

balanced?

Explanation:

because if it wasn't moving that means they are pulling at a similar strength

patriot [66]2 years ago

3 0

Either balanced or at rest.

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When an object slides across the floor, it is slowed down by friction and the floor surface gets warmer. what energy conversion

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2 years ago

A 62.0 kgkg skier is moving at 6.90 m/sm/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch

Margarita [4]

Answer:

the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

the internal energy generated in crossing the rough patch is 820.26 J

Explanation:

Given that,

Mass, m = 62 kg,

Initial speed, = 6.90 m/s

Length of rough patch, L = 4.50 m,

coefficient of friction, = 0.3

Height of inclined plane, h = 2.50 m

$g=9.8 m/s^2$ is the acceleration due to gravity

According to energy conservation equation,

Part (b)

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

$F_f=\mu mg$

Therefore, the work done by friction is:

$W=-F_f d =-\mu mg d$

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

$W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J$

So, the internal energy generated in crossing the rough patch is 820.26 J

Part (a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

$E=K_i +U_i = \frac{1}{2}mu^2 + mgh$

After crossing the rough patch, the new mechanical energy is

$E'=E+W$

where

W = -820.26 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

$E' = K_f = \frac{1}{2}mv^2$

where v is the final speed.

According to the law of conservation of energy, we can write:

$E+W=E'$

So we find v:

$\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s$

Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

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2 years ago

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