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3 years ago

What kind of friction occurs when moving parts have ball bearings

Physics

2 answers:

Oksanka [162]3 years ago

7 0

Answer: Rolling friction

Explanation:

Rolling friction is said to act when an object rolls over a surface. Machines often have moving parts. These moving parts roll over each other as the machine works. Ball bearings are commonly used to reduce friction between the two surfaces in contact as the machine parts roll over each other. This is a typical example of rolling friction as applied to machine parts.

ASHA 777 [7]3 years ago

7 0

Answer:

Rolling friction.

Explanation:

Rolling friction is the force resisting the motion when a body ( such as a ball, tires, wheel) rolls on a surface. It is applicable for bodies whose point of contact keeps changing. It is the force that opposes the motion of a body which is rolling over the surface of another.

Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation of the wheel, roadbed, etc. is recovered when the pressure is removed.

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Answer:

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Explanation:

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2 years ago

A battery operated DVD player uses 12 DV volts from AA batteries and draws a current of 3 A. Calculate the resistance.

serg [7]

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3 years ago

A man claims that he can hold onto a 16.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

Anna007 [38]

Answer:

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

Explanation:

Given;

Mass of child m = 16 kg

Speed of each car v = 59.0 mi/h = 26.37536 m/s

Time t = 0.05s

Applying the impulse momentum equation;

Impulse = change in momentum

Ft = ∆(mv)

F = ∆(mv)/t

F = m(∆v)/t

Where;

F = force

t = time

m = mass

v = velocity

Since the final speed of the car is zero(at rest) then;

∆v = 0 - v = -26.37536 m/s

Substituting the given values;

F = 16×-26.37536/0.05

F = -8440.1152 N

F = -8440.12 N

the magnitude of the average force needed to hold onto the child is 8440.12 N

5 0

3 years ago

Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022

monitta

Answer:

$F=1.95\times 10^{20}\ N$

Explanation:

Mass of Earth, $m_e=5.98 \times 10^{24}\ kg$

Mass of Moon, $m_m=7.26\times 10^{22}\ kg$

The distance between Earth and the Moon is, $d=384,400\ km$

We need to find the force of gravitational attraction between the Earth and the moon. The force of gravity is given by :

$F=G\dfrac{m_em_m}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{5.98 \times 10^{24}\times 7.26\times 10^{22}}{(384400 \times 10^3)^2}\\\\F=1.95\times 10^{20}\ N$

So, the required force is $1.95\times 10^{20}\ N$ .

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2 years ago

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, A)compression B)w

Gnesinka [82]

Salutations!

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, _______________ is being done.

<span>If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done. Energy being transferred and the object begins to move is called work.

Thus, your answer is option B.

Hope I helped (:

Have a great day!</span>

7 0

3 years ago

Read 2 more answers