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TEA [102]
3 years ago
13

What kind of situation can you think would be a benefit of reducing friction?

Physics
1 answer:
zepelin [54]3 years ago
8 0

Answer:

Wear and tear of an object is not desirable as it reduces its life. This is more so in case of moving parts in automobiles and machinery. Therefore, efforts are made to reduce friction between moving parts. Friction between moving parts is usually reduced by introducing a substance between the moving surfaces.

Explanation:

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An object is moving to the west at a constant speed. three forces are exerted on the object. one force is 10 n directed due nort
SVETLANKA909090 [29]
If we want the object to continue to move at constant speed, it means that the resultant of the forces acting on the object must be zero. So far, we have:
- force F1 with direction north, of 10 N
- force F2 with direction west, of 10 N
The third force must balance them, in order to have a net force of zero on the object.

The resultant of the two forces F1 and F2 is
F_{12} =  \sqrt{F_1^2+F_2^2}= \sqrt{(10 N)^2+(10 N)^2}= \sqrt{200}=14.1 N
with direction at 45^{\circ} north-west. This means that F3 must be equal and opposite to this force: so, F3 must have magnitude 14.1 N and its direction should be 45^{\circ} south-east.
6 0
3 years ago
(I will give brainliest whoever helps me !!)
lorasvet [3.4K]
C. Forces have mass and take up space
3 0
2 years ago
A block of ice of mass 4.30 kg is placed against a horizontal spring that has a force constant k = 250 N/m and is compressed a d
OleMash [197]

Answer:

W = 0.060 J

v_2 = 0.18 m/s

Explanation:

solution:

for the spring:

W = 1/2*k*x_1^2 - 1/2*k*x_2^2

x_1 = -0.025 m and x_2 = 0

W = 1/2*k*x_1^2 = 1/2*(250 N/m)(-0.028m)^2

W = 0.060 J

the work-energy theorem,

W_tot = K_2 - K_1 = ΔK

with K = 1/2*m*v^2

v_2 = √2*W/m

v_2 = 0.18 m/s

8 0
3 years ago
Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal
Dmitry_Shevchenko [17]

Answer:

a = \dfrac{4\pi^2R}{T^2}

Explanation:

The acceleration of a circular motion is given by

a = \omega^2 R

where \omega is the angular velocity and R is the radius.

Angular velocity is related to the period, T, by

\omega=\dfrac{2\pi}{T}

Substitute into the previous formula.

a = (\dfrac{2\pi}{T})^2 R

a = \dfrac{4\pi^2R}{T^2}

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0
2 years ago
A satellite is revolving the earth 4km above the surface.find the orbital velocity of the satellite (R =6400km,g=9.8m/s^2)​
Karo-lina-s [1.5K]

Answer:

v ≈ 7900 m/s

Explanation:

centripetal force will equal gravity force

mv²/R = mg

v²/R = g

v² = Rg

v = √(Rg)

v = √(6.4e6(9.8))

v = 7.91959...e+3

v ≈ 7900 m/s

of course, at those velocities and that deep into the atmosphere, the satellite would quickly burn up, slow down, and cause tremendous damage to buildings etc. with the sonic boom shock wave. It would also have to avoid a lot of mountains as 4000 m is not that high.

3 0
3 years ago
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