Answer:
The magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force. ( ¹/₄ F)
Explanation:
The electrostatic force between two charges is given by Coulomb's law;
where;
Q₁ and Q₂ are the magnitude of the charges
r is the distance between the charges
k is Coulomb's constant
Since the charges are identical;
Q₁ = Q
Q₂ = Q
the electrostatic force experienced by each charge is given by;
When each of the spheres has lost half of its initial charge;
Q₁ = Q/2
Q₂ = Q/2
Therefore, the magnitude of electrostatic force on each charge is quarter of the magnitude of initial electrostatic force.
The hand saw involved more work and the electric saw involved more power.
Answer:
the answer to the question is calcium
D = m /v
d = 30 / 0.5 Kg/m³
d = 60 Kg / m³
In short, Your Answer would be Option A
Hope this helps!
<u><em>Correct question:</em></u><u><em>a </em></u><u><em>200 g </em></u><u><em>ball is dropped from a height of 2m, bounces on a hard floor and rebounds to a height of 1.5m. What maximum force does the floor exert on the ball?</em></u>
<u><em>The diagram of the question is in the attachment.</em></u>
Answer:
Explanation:
V=√2gh
g-10m/s²
let u be initial velocity ad v be final velocity,
u=√2*10*2
u=6.324m/s
v=√2*10*1.5
v=5.477m/s
from the diagram
t=5ms=0.005s
F=-33.87N (the negative shows direction)
From the diagram Fmax=2F
Fmax= 2*33.87
=67.74N
