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2 years ago

An object, experiencing no friction, keeps moving at a constant speed. What can we say about the net force on the object?

Physics

1 answer:

sleet_krkn [62]2 years ago

3 0

Answer:

Explanation:

Let's look at a mathematical representation of this. The equation for tis is just a souped up version of Newton's 2nd Law:

F - f = ma. It an object is moving at a constant speed, the acceleration of that object is 0. That changes this equation to

F = f which states that the applied Force equals the frictional force, choice a.

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It has been suggested that rotating cylinders about 14.5 mi long and 4.78 mi in diameter be placed in space and used as colonies

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Angular speed must such a cylinder have so that the centripetal acceleration at its surface equals the free-fall acceleration on Earth is 0.0466 rad/s.

Length of the cylinder, L = 9 mi

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v = r ω

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6 months ago

During takeoff, an airplane goes from 0 to 56 m/s in 9 s. How fast is it going after 4 s?

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3 years ago

A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.

ikadub [295]

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) $8.75\ m/sec^2$

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

$Fh_{net}=F-F_r=50 N - 15N=35N$

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

$F_n=35N$ in the x-axis

e) Following Newton's second law, the acceleration is calculated as

$a=\frac{F_t}{m}=\frac{35N}{4Kg}=8.75m/sec^2$

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