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Serggg [28]
3 years ago
6

A 70-kg boy is sitting 3 m from the ground in a tree. What is his gravitational potential energy

Physics
1 answer:
steposvetlana [31]3 years ago
7 0

m x h x 9.8 m/s squared

70 kg x 3 m x 9.8 m/s squared= 2058 Joules

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Exons And Introns differ because Exons code for protein and Introns do not.

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Someone please help me...
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Which wave must have a medium to travel?
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6 0
3 years ago
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A 1 pound bucket rests on a table. What is the support force exerted on the bucket by the table? What is the support force when
katrin2010 [14]

This question involves the concepts of equilibrium and Newton's third law of motion.

The support force will be "1 pound" for the empty bucket and the support force will be "6 pounds" after pouring water into it.

  • According to the condition of equilibrium, the sum of forces acting on a stationary object must be zero. Hence, the support force of the table will be equal to the total mass of the bucket.
  • According to Newton's Third Law of Motion every action force has an equal but opposite reaction force. Hence, the support force will be a reaction force to the weight of the bucket.

Therefore, the support force in each case will be equal to the total mass of the bucket:

Case 1 (empty bucket):

<u>support force = 1 pound</u>

<u></u>

Case 1 (water poured):

support force = 1 pound + 5 pound

<u>support force = 6 pound</u>

<u></u>

Learn more about equilibrium here:

brainly.com/question/9076091

8 0
3 years ago
A 2.00 kg box slides on a rough, horizontal surface, hits a spring with a speed of 1.90 m/s and compresses it a distance of 10.0
oksian1 [2.3K]

Answer:

Explanation:

Given

mass of box m=2\ kg

speed of box v=1.9\ m/s

distance moved by the box x=10\ cm

coefficient of kinetic friction \mu _k=0.66

Friction  force f_r=\mu_kN

f_r=0.66\times mg

f_r=0.66\times 2\times 9.8=12.936 \N

Kinetic Energy of box will be utilize to overcome friction and rest is stored in spring in the form of elastic potential energy

\frac{1}{2}mv^2=f_r\cdot x+\frac{1}{2}kx^2

\frac{1}{2}\times 2\times 1.9^2=12.936\times 0.1+\frac{1}{2}\times k\times (0.1)^2

3.61-1.2936=0.005\times k

k=463.28\ N/m

3 0
3 years ago
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