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stellarik [79]
3 years ago
11

A block of mass 9.5kg rests on a slope an angle of 23.0∘ relative to the horizontal. What is the size of the contact force norma

l to the slope, in Newtons?
Use a gravitational acceleration value of g = 9.8m/s^2.
Physics
1 answer:
Rudiy273 years ago
6 0

The Normal Force = M x G x Cos(theta)

= 9.5 Kg x 9.8 m/s^2 x cos 23

= 9.5 Kg x 9.8 m/s^2 x 0.9205

Converting Kg to Newton,

1 Kg  = 9.81 N

= 9.5 Kg x 9.81 N x 9.8 m/s^2 x 0.9205

= 840.702 N

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_________________

v = r*omega

omega = (delta thaeda)/(delta time)

_________________

2.5 m/s = 30m*omega

.08333...  = omega

_________________

*note, omega stands for angular velocity

__________________

Also, it's been a while since I've done a physics problem like this so I could be wrong. I think this is the answer though :)

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Answer:

________________

<em><u>560</u></em><em><u> </u></em><em><u>kilometers</u></em><em><u> </u></em>

I think this is correct.

________________

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3 years ago
Read 2 more answers
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