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zhenek [66]
3 years ago
6

Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp

ose) and throw a rock. What will the rock do? Why?​
Physics
1 answer:
xz_007 [3.2K]3 years ago
6 0
The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.
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A gas has a pressure of 220 kPa at a volume of 380K. At what temperature will the gas have a pressure of 256 kPa?
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The temperature will be the pressure of 130 ka
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An electric grinder uses a grinding wheel
luda_lava [24]
(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s² 

<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>

<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
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3 years ago
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Vitek1552 [10]

Answer:

A.

Explanation:

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A uniform 1.6-kg rod that is 0.89 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs han
artcher [175]

Answer:

8.27°

Explanation:

To angle difference will be determined by the difference in the displacement of the springs, produced by the weight of the center of mass of the rod.

d=y_1-y_2=\frac{F_1}{k_1}-\frac{F_2}{k_2}=\frac{0.5mg}{31N/m}-\frac{0.5mg}{63N/m}\\\\d=0.5(1.6kg)(9.8m/s^2)[\frac{1}{31N/m}-\frac{1}{63N/m}]=0.128m

by a simple trigonometric relation you obtain that the angle:

sin\theta=\frac{d}{l}=\frac{0.128m}{0.89m}=0.144\\\\\theta=sin^{-1}(0.144)=8.27\°

hence, the angle between the rod and the horizontal is 8.27°

4 0
3 years ago
An asteroid is moving along a straight line. A force acts along the displacement of the asteroid and slows it down. The asteroid
Setler79 [48]

Answer:

a)

- 7.04\times10^{11} J

b)

5.03\times10^{5} N

Explanation:

a)

m = mass of the asteroid = 43000 kg

v_{o} = initial speed of asteroid = 7600 m/s

v = final speed of asteroid = 5000 m/s

W = Work done by the force on asteroid

Using work-change in kinetic energy theorem

W = (0.5) m (v^{2} - v_{o}^{2})\\W = (0.5) (43000) (5000^{2} - 7600^{2})\\W = - 7.04\times10^{11} J

b)

F = magnitude of force on asteroid

d = distance traveled by asteroid while it slows down = 1.4 x 10⁶ m

Work done by the force on the asteroid to slow it down is given as

W = - F d\\- 7.04\times10^{11} = - F (1.4\times10^{6})\\F = 5.03\times10^{5} N

4 0
3 years ago
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