How could you increase the gravitational potential energy of an object without changing its mass and gravity?A. Make the object larger
B. Raise the object farther off the ground
C. Lower the object towards the ground
D. Allow the object to roll on the ground
Answers
Answer:
B Raise the object farther off the ground
Final answer:
To increase the gravitational potential energy of an object without altering its mass or gravity, you would need to raise the object to a greater height. The potential energy is determined by the object's mass, its height, and the gravitational force, as shown by the formula potential energy=mgh.
Explanation:
To increase the gravitational potential energy of an object without changing its mass and gravity, you could raise the object farther off the ground. This is because gravitational potential energy is a function of an object's mass, height, and acceleration due to gravity, as represented by the formula potential energy = mgh, m being mass, g being gravity, and h being height.
Moving the object to a greater height without acceleration or carrying the object with or without acceleration at the same height will not result in an increased potential energy. Only by raising the object to a higher position or altitude, you increase its potential energy.
In essence, the principle involves work done against the gravity. When an object is raised to a higher elevation, work is done against gravity. This work gets stored as potential energy in the object-Earth system. Thus, the higher the position of the object, the higher would be its gravitational potential energy.
Learn more about Gravitational Potential Energy here:
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Choose the best answer for each situation. A scientist with the EPA examining water for contamination writes down that there was an iridescent sheen on the surface of the water. What type of data did the scientist record?
Answers
Answer:
The answer is 2.71 g/mL
Explanation:
The density of a substance can be found by using the formula
From the question
mass = 244 g
volume = 90 mL
So we have
We have the final answer as
2.71 g/mL
Hope this helps you
Answers
Answers
Answer:
The answer of this question is =1.258*10-4
Answers
Vi=0m/s
Vf=143.8m/s
A=-9.8m/s
d=???
Use the equation Vf^2=Vi^2+2A(d)
Rearrange to isolate d: d=Vf^2/2A
d=(143.8)^2/2(-9.8)
d=20678.4/-19.6
d=-1055m
The tank was released from a height of 1055m
period of 1.262 Earth days and an orbital radius of 2.346 x 104
km. The second moon has an orbital
radius of 9.378 x 103 km.
What is the orbital period of the second moon?
Answers
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.
Moon #1: (1.262 days)² / (2.346 x 10^4 km)³
Moon #2: (orbital period)² / (9.378 x 10^3 km)³
If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.
Cross multiply the proportion:
(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³
Divide each side by (2.346 x 10^4)³:
(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³
= 0.1017 day²
Orbital period = 0.319 Earth day = about 7.6 hours.
Answers
Answer
33,000,000 joules or 33MJ
Explanation:
The formula for calculating the heat released by a substance is expressed as shown below;
where;
Q is the amount of heat released
p is the power expended
t is the time it takes
Then;
Given the current I = 100A, supply voltage V = 220V and required time t = 25mins
t = 25 * 60 = 1500 seconds
Substituting these values in the formula that we have;