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Elan Coil [88]
3 years ago
9

A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti

ng on the block? Round your answer to the nearest whole number.
Physics
1 answer:
aleksley [76]3 years ago
7 0

Magnitude of normal force acting on the block is 7 N

Explanation:

10N = 1.02kg

Mass of the block = m = 1.02 kg

Angle of incline Θ =  30°

Normal force acting on the block = N

From the free body diagram,

N = mgCos Θ

N = (1.02)(9.81)Cos(30)

N = 8.66 N

Rounding off to nearest whole number,

N = 7 N

Magnitude of normal force acting on the block = 7 N

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A 64.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxyg
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3 years ago
The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potentia
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Answer:

Energy Lost for group A's car = 0.687 J

Energy Lost for group B's car = 0.55 J

Explanation:

The exact question is as follows :

Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.

To find - How much energy is lost due to heat for group A's car ?

              How much for Group B's car ?

Solution -

We know that,

GPE = 1 Joule (Potential Energy)

Now,

For Group A -

Energy Lost = GPE - KE

                    = 1 J - 0.313 J

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So,

Energy Lost for group A's car = 0.687 J

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For Group B -

Energy Lost = GPE - KE

                    = 1 J - 0.45 J

                    = 0.55 J

So,

Energy Lost for group B's car = 0.55 J

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On average, how many stars would we have to search before we would expect to hear a signal? Assume there are 500 billion stars in the galaxy.

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