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exis [7]
3 years ago
12

Which of the following objects have gravitational potential energy?

Physics
2 answers:
Jlenok [28]3 years ago
6 0

Answer:

an apple on a tree branch 3m above the ground

Explanation:

Brut [27]3 years ago
3 0

Answer:

The apple

Explanation:

The apple has gravitational potential energy because it is just sitting there but nothing is pushing it up, it is not sitting on something. This means that it has a lot of gravitational potential energy as nothing is pushing it up.

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Please help 9.2.1 project in science just ned an example​
olga55 [171]

Answer:

Give me what kind of example you need please so I can help you. Put it in the comments.

Explanation:

3 0
3 years ago
A pair of electrically charged objects attract each other with a force of 4 N when they are a distance of 3 m apart. If their ch
Rainbow [258]

Answer:

1 N

Explanation:

From coulomb's law,

The force of attraction between two charges is inversely proportional to the square of the distance between the charges.

From the question,

Assuming the charges are the same in both case,

F ∝ /r²....................... Equation 1

Fr² = k

F'r'² = Fr²........................... Equation 2

Where F' = First Force, r'² = First distance, F = second force, r² = second distance.

make F the subject of the equation,

F = F'r'²/r².................... Equation 3

Given: F' = 4 N, r' = 3 m, r = 6 m

Substitute into equation 3

F = 4(3²)/6²

F = 36/36

F = 1 N

7 0
3 years ago
Give some example acostic using "ELEMENTS"
Jet001 [13]
E=ERBIUM
L=LITHIUM
E=EINSTEINIUM
M=MAGNESIUM
E=EUROPIUM
N=NITROGEN
T=TITANIUM
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5 0
3 years ago
Can someone explain please <br> ???
Radda [10]

Answer: A

Explanation:

6 0
3 years ago
A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid
Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm=\frac{6}{100}=0.06 m

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

dF=IBdr

dF=IBdr

\tau=rdF=IBrdr

\tau=\int_{0}^{R}IBr dr

\tau=IB(\frac{R^2}{2}

Torque due to friction

\tau=R\times F

Where friction force=F

R\times F=\frac{IBR^2}{2}

F=\frac{IBR}{2}

Substitute the values

F=\frac{3\times 1\times 0.06}{2}

F=0.09 N

7 0
3 years ago
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