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Alchen [17]
4 years ago
15

What kind of friction occurs when moving parts have ball bearings

Physics
2 answers:
Oksanka [162]4 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]4 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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A jet is circling an airport control tower at a distance of 20.6 km. An observer in the tower watches the jet cross in front of
lesya [120]

Answer:

197.76 m

Explanation:

r = Radius of the path = 20.6 km = 20.6\times 10^3\ m

\theta = The angle subtended by moon = 9.6\times 10^{-3}\ rad

Distance traveled is given by

s=r\times\theta

\Rightarrow s=20.6\times 10^3\times 9.6\times 10^{-3}

\Rightarrow s=197.76\ m

The distance traveled by the jet is 197.76 m

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3 years ago
Will mark BRAINLIEST!!!!
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3 years ago
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dem82 [27]
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3 years ago
Two moles of an ideal monatomic gas are at a temperature of 345 K. Then, 2250 J of heat is added to the gas, and 870 J of work i
Lapatulllka [165]

Answer: The final temperature is 470K

Explanation: Using the relation;

Q= ΔU +W

Given, n = 2mol

Initial temperature T1= 345K

Heat =Q= 2250J

Workdone=W=-870J(work is done on gas)

T2 =Final temperature =?

ΔU =3/2nR(T2-T1)

ΔU=3/2 × 2 ×8.314 (T2 - 345)

ΔU=24.942(T2-345)

Therefore Q = 24.942(T2-345)+ (-870)

2250=24.942(T2-345)+ (-870)

125.09=(T2-345)

T2 =470K

Therfore the final temperature is 470K

5 0
3 years ago
Two forces act on a moving object that has a mass of 27 kg. One has a magnitude of 12 N and points due south, while the other ha
levacccp [35]

0.77 m/s2 directed 35° south of west

net force = (-17,-12)

net force = mass * acceleration

(-17,-12) = 27 * (x-acceleration,y-acceleration)

(x-acceleration,y-acceleration) = (-17/27,-12/27) = (-0.629629629..., -0.444...)

angle of acceleration = tan^-1 (-0.444.../-0.629629...) = 35.21759 degrees below negative x-axis.

magnitude of acceleration = sqrt((-0.629629...)^2 + (-0.444...)^2) = 0.77069 (5dp)

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3 years ago
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