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adoni [48]
3 years ago
7

When a person holds a ball above earth's surface, where is this potential energy stored?​

Physics
1 answer:
inn [45]3 years ago
3 0

Answer:

in the ball

Explanation:

potential energy is where an object has energy, always, but it isn't always moving, and when you drop the ball, it has kinetic energy.

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PLS HELP!!!- The moon rotates around the Earth every 28 days. What is the frequency of the moon’s revolution?
Naily [24]

Answer: the frequency is every 27.322 days

6 0
3 years ago
How is the electrostatic force affected when the magnitude of a charge is doubled?
BaLLatris [955]
The magnitude of the electrostatic force between two charges is given by:
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We can see that the magnitude of the force is directly proportional to the charges. This means that when one of the charges is doubled, the magnitude of the electrostatic force will double as well, so the correct answer is
A) <span>The magnitude of the electrostatic force doubles</span>
4 0
3 years ago
Hey pls help thanks a lot
galben [10]

Answer:

I am not sure about the answer as I don't have a proper calculator besides me now

Explanation:

but I used this equation:

(8.20)sin30(1-d)=10d

Idk whether it is correct or not, I'm just a student too

what is your method of doing this question?

4 0
3 years ago
An object propelled upwards with an acceleration of 2.0 m / s ^ 2 is launched from rest. After 6 seconds the fuel runs out. Dete
dezoksy [38]

Answer:43.34 m

Explanation:

Given

acceleration(a)=2 m/s^2

Initial Velocity(u)=0 m/s

After 6 s fuel runs out

Velocity after 6 s

v=u+at

v=0+2\times 6=12 m/s

After this object will start moving under gravity

height reached in first 6 s

s=ut+\frac{at^2}{2}

s=0+\frac{2\times 6^2}{2}

s=36 m

After fuel run out distance traveled in upward direction is

v^2-u^2=2as_0

here v=0

u=12 m/s

a=9.8 m/s^2

0-12^2=2(-9.8)(s)

s_0=\frac{144}{2\times 9.8}=7.34 m

s+s_0=36+7.34=43.34 m

7 0
3 years ago
A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell
Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem  we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0
3 years ago
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