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Lana71
3 months ago
9

If Pete ( mass=90.0kg) weights himself and finds that he weighs 30.0 pounds, how far away from the surface of the earth is he

Physics
1 answer:
serg [3.5K]3 months ago
5 0
The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.
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The spring in a retractable ballpoint pen is 1.8 cm long, with a 300 N/m spring constant. When the pen is retracted, the spring
Sav [3153]

Answer:

The pen requires 7.2 mJ of energy to extend.

Explanation:

Provided:

Length = 1.8 cm

Spring constant = 300 N/m

Initial compression = 1.0 mm

Additional compression = 6.0 mm

Total compression = 1.0 + 6.0 = 7.0 mm

We need to determine the energy needed

This energy is equivalent to the variation in spring potential energy

E=PE_{2}-PE_{1}

E=\dfrac{1}{2}kx_{2}^2-\dfrac{1}{2}kx_{1}^2

Substitute the values into the formula

E=\dfrac{1}{2}\times300\times(7.0\times10^{-3})^2-\dfrac{1}{2}\times300\times(1.0\times10^{-3})^2

E=0.0072\ J

E=7.2\ mJ

Therefore, a total of 7.2 mJ is needed to extend the pen.

7 0
4 months ago
Read 2 more answers
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