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snow_lady [41]
3 years ago
15

What is the potential energy of a 32 kg child sitting on the floor? Explain.

Physics
1 answer:
Rama09 [41]3 years ago
3 0
Potential energy= mass x gravitational field x height
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8TH GRADE SCIENCE I NEED NOW DO NOT SKIP
yan [13]

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

=50(7.8)= 390N

7 0
3 years ago
What are some examples of gravitational forces that keep an object in orbit? Pls give at least 2-3 example if you can.​
Lelu [443]

Answer:

Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.

1. satellite in orbit around the earth             (motion of earth is negligible)

2. moon in orbit around the earth     (center of motion several thousand miles        

                                                             from center of earth)

3. earth in orbit around sun    (center of rotation close to center of sun)

4. binary stars         (if masses of stars are equal center of rotation is in middle)

3 0
3 years ago
Describe in your own words what is meant by the statement that the Sun is in hydrostatic equilibrium.
Jet001 [13]

Answer and Explanation:

Hydrostatic equilibrium is the condition in which force is balance that is upward force and downward force the downward force is due to gravitational force and the upward force is due to the pressure. The Sun is said to be in hydrostatic equilibrium means the force acting on it is balance means upward force which is due to pressure is same as the force exerted by gravitation.

8 0
4 years ago
Can anyone help? Due tomorrow!!
zvonat [6]
Different:
1. number of moons
2. size of rings
3. mass
4. temperature
5. type of rings

i might be wrong though.
7 0
3 years ago
A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose d
Lilit [14]

Answer:

the tension in the part of the cord attached to the textbook is 7.4989 N

Explanation:

Given the data in the question;

As illustrated in the image below;

first we determine the value of the acceleration,

along vertical direction; we use the second equation of motion;

y = ut + \frac{1}{2}a_yt²

we substitute;

0 m/s for u, 1.29 m for y, 0.850 s for t,

1.29 = 0×0.850  + \frac{1}{2}×a_y×(0.850)²

1.29 = 0.36125a_y

a_y = 1.29 / 0.36125

a_y = 3.5709 m/s²

Now when the text book is moving with acceleration , the dynamic equation will be;

T₁ = m₁a_y

where m₁ is the mass of the text book ( 2.10 kg )

a_y is the vertical acceleration ( 3.5709 m/s² )

so we substitute

T₁ = 2.10 × 3.5709

T₁ = 7.4989 N

Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N

3 0
3 years ago
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