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3 years ago

The amount of force needed to sustain motion of a rock in outer space is

Physics

1 answer:

BartSMP [9]3 years ago

4 0

Answer:

0 N

Explanation:

According to Newton's first law, object subjected to no force or 0 net force will stay in the same motion. In outer space, there's no gravity or air resistance. So if the net force is 0, a rock in motion would be still in motion forever without a need of an extra force to sustain it.

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A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

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How to calculate the uncertainty

agasfer [191]

Answer:

If you're taking the power of a number with an uncertainty, you multiply the relative uncertainty by the number in the power.

Explanation:

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_____ is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.

Juli2301 [7.4K]

Rotational kinetic energy <span>is the kinetic energy of an object, proportional to the object's moment of inertia and the square of its angular velocity.</span>

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Velocity is a vector quantity which has both magnitude and direction. Using complete sentences, describe the object’s velocity d

andreyandreev [35.5K]

Answer:

Explanation:

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One Newton equals .225

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The answer to your question is 10.24

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