The total charge on the interior of the conductor is zero.
The total charge on the exterior of the conductor is 8q.
<h3>
Total charge on the interior</h3>
Due to large number of electrons available for conduction in a conductor, most of the electrons moves to surface leaving zero net charge inside the conductor.
Thus, the total charge on the interior of the conductor is zero.
<h3>T
otal charge on the exterior</h3>
The total charge on the exterior of the conductor is calculated as follows;
Q = q + 7q = 8q
Thus, the total charge on the exterior of the conductor is 8q.
Learn more about net charge on interior and exterior of conductors here: brainly.com/question/14653264
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Answer:
The ball will fall on the X .
Explanation:
At height, when the aeroplane is in great speed , everything attached with it acquires the same speed . So ball will also have the same speed as the aeroplane have. When ball starts falling off , it gets detached from plane but , at the same time it continues to travel with its earlier speed , because of inertia of motion. So it remains stationary with respect to plane in horizontal direction . It has velocity with respect to plane only in vertical direction. Hence it will fall on the X. It is due to first law of motion.
Answer:
v1 = 15.90 m/s
v2 = 8.46 m/s
mechanical energy before collision = 32.4 J
mechanical energy after collision = 32.433 J
Explanation:
given data
mass m = 0.2 kg
speed = 18 m/s
angle = 28°
to find out
final velocity and mechanical energy both before and after the collision
solution
we know that conservation of momentum remain same so in x direction
mv = mv1 cosθ + mv2cosθ
put here value
0.2(18) = 0.2 v1 cos(28) + 0.2 v2 cos(90-28)
3.6 = 0.1765 V1 + 0.09389 v2 ................1
and
in y axis
mv = mv1 sinθ - mv2sinθ
0 = 0.2 v1 sin28 - 0.2 v2 sin(90-28)
0 = 0.09389 v1 - 0.1768 v2 .......................2
from equation 1 and 2
v1 = 15.90 m/s
v2 = 8.46 m/s
so
mechanical energy before collision = 1/2 mv1² + 1/2 mv2²
mechanical energy before collision = 1/2 (0.2)(18)² + 0
mechanical energy before collision = 32.4 J
and
mechanical energy after collision = 1/2 (0.2)(15.90)² + 1/2 (0.2)(8.46)²
mechanical energy after collision = 32.433 J
