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2 years ago

A sled sliding on a flat,icy surface with a constant velocity is best described by

1 answer:

5 0

The first one is actually Newton's First Law of Motion.

Newton's First Law of Motion is commonly stated as "an object in motion will remain in motion unless acted upon by an outside force." This sled is moving across the ice at the same constant speed and velocity meaning it will won't stop unless an outside force acts upon it. An example being someone coming and grabbing the sled or maybe the sled slips off of the ice and finds itself stuck in a pile of snow. Similarly, if the sled was not moving, it would not move unless acted upon again by an outside force.

You believed it to be Newton's Third Law, but for future reference, his third law is more about every action having an equal opposite reaction. This wouldn't explain what is currently happening with the sled. A helpful tip into figuring out where to start in figuring out the answer I got is in the example you gave it literally states "with a constant velocity" which also ends up being in the description of Newton's First Law. Though, don't rely on that because the answer could have had a made up definition.

Unfortunately, I do not know the answer to your other question. I apologize. Hope this helped.

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

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2 years ago

If the 5-N force and the 12-N force form a 90 degree angle, what is the magnitude of the force acting in the direction of the da

Leni [432]

Answer:

Explanation:

<em>Kindly see attached file for your reference</em>

Step one:

given data

the horizontal component of the force= 12N

the vertical component of the force= 5N

The dashed arrow represents the hypotenuse of the triangle, hence the resultant of the force system.

By implication of this, we will use the Pythagoras theorem to solve for the resultant force

Step two:

$F_R=\sqrt{F_H^2+F_V^2}\\\\F_R= \sqrt{12^2+5^2}\\\\F_R=\sqrt{144+25}\\\\F_R=\sqrt{169}\\\\F_R=13N$

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Answer:

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Explanation:

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2 years ago

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A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begin

Mashutka [201]

The given question is incomplete. The complete question is as follows.

A 75-g bullet is fired from a rifle having a barrel 0.540 m long. Choose the origin to be at the location where the bullet begins to move. Then the force (in newtons) exerted by the expanding gas on the bullet is $14,000 + 10,000x − 26,000x^{2}$ , where x is in meters. Determine the work done by the gas on the bullet as the bullet travels the length of the barrel.