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o-na [289]
3 years ago
8

If you exert a horizontal force of 200 N to slide a crate across a factory floor at constant velocity, how much friction is exer

ted by the floor on the crate?
Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

200 N

Explanation:

Applied force, F = 200 N

Let the friction force is f.

By use of Newton's second law

Net force = mass × acceleration

F - f = m × a

Here acceleration a is zero as crate moves with constant velocity.

So, F - f = 0

f = F

Friction force = 200 N

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Energy in many of its forms may be used in natural processes, or to provide some service to society such as heating, refrigeration, lighting or performing mechanical work to operate machines. For example, in order to heat your home, your furnace can burn fuel, whose chemical potential energy is thus converted into thermal energy, which is then transferred to your home's air in order to raise its temperature.


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3 0
3 years ago
Read 2 more answers
Help !!
777dan777 [17]
<span>One leg is  = 12 m, and the other leg is 16 m. </span>
3 0
3 years ago
Find the hiker’s gravitational potential energy if the cliff is 60m high
Furkat [3]

Answer:

Potential energy is U=mgh

Explanation:

The potential energy depends on the mass, the acceleration of gravity g and the height at which the object or person is.

Potential energy  U=mgh

In this case we would need to know the exact mass of the hiker in order to calculate the potential energy.

But we know the values of g and h

g=9.81m/s^2

h=60m

So, the potential energy

U=m(9.81m/s^2)(60m)\\\\U=588.6*m

m is the mass of the hiker, wich is not in the description of the problem.

4 0
3 years ago
A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum outp
Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H  =  71 m

Acceleration due to gravity g =9.8 m/s^2

density of water ρ= 1000 kg/m^3

The minimum output gauge pressure to make water reach height H

P= ρgH

= 1000×9.8×71= 695800 N/m^2 or Pa

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3 years ago
At a particular instant the magnitude of the momentum of a planet is 2.05 × 10^29 kg·m/s, and the force exerted on it by the sta
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