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3 years ago

Garrick rubs an inflated balloon against his hair. He then touches the balloon against a non-conducting wall.

Physics

1 answer:

Sauron [17]3 years ago

7 0

Answer:

Figure A

Explanation:

At first, the inflated balloon is rubbed against the hair.

In this situation, the balloon is charged by friction: because of the friction between the surface of the balllon and the hair, electrons are transferred from the hair to the surface of the balloon.

As a result, when the balloon is detached from the hair, it will have an excess of negative charge (due to the acquired electrons).

Then, the balloon is placed in contact with the non-conducting wall.

The non-conducting wall is initially neutral (equal number of positive and negative charges).

Because the wall is made of a non-conducting material (=isolant), the charges cannot move easily through it. Therefore, even though the charges on the wall feel a force due to the presence of the electrons in the balloon, they will not redistribute along the wall.

Therefore, the charges on the wall will remain equally distributed, as shown in figure A.

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6. The hole on a level, elevated golf green is a horizontal distance of 150 m from the tee and at an elevation of 12.4 m above t

Georgia [21]

Answer:

u = 104.68 m/s

Explanation:

given,

horizontal distance = 150 m

elevation of 12.4 m

angle = 8.6°

horizontal motion = x = u cos θ. t .............(1)

vertical motion =

$y = u sin \theta - \dfrac{1}{2}gt^2$ ................(2)

from equation(1) and (2)

$y = x tan \theta - \dfrac{gx^2}{2u^2cos^2\theta}$ ..........{3}

$12.4 = 150\times tan (8.6) - \dfrac{9.8\times 150^2}{2u^2cos^2(8.6)}$

$\dfrac{9.8\times 150^2}{2u^2cos^2(8.6)} = 10.29$

$\dfrac{9.8\times 150^2}{2\times 10.29\times cos^2(8.6)} = u^2$

$u = \sqrt{10959.34}$

u = 104.68 m/s

The initial speed of the ball is u = 104.68 m/s

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3 years ago

Car a runs a red light and broadsides car b, which is stopped and waiting to make a left turn. car a has a mass of 1,800kg. car

frez [133]

The law of conservation of momentum tells us that momentum is conserved, therefore total initial momentum should be equal to total final momentum. In this case, we can expressed this mathematically as:

mA vA + mB vB = m v

where, m is the mass in kg, v is the velocity in m/s

since m is the total mass, m = mA + mB, we can write the equation as:

mA vA + mB vB = (mA + mB) v

furthermore, car B was at a stop signal therefore vB = 0, hence

mA vA + 0 = (mA + mB) v

1800 (vA) = (1800 + 1500) (7.1 m/s)

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