a) An inflated balloon was pressed against a wall after it has been rubbed with a piece of synthetic cloth. It was found that the balloon sticks to the wall. This is because a positive and negative electric charge is produced, therefore the balloon sticks to the wall.

b) When an object is thrown up, it comes back to ground because of gravitational attraction force of earth.

c) Mountaineers suffer nose bleeding at higher altitudes because the oxygen level decreases with increase in altitude, which the body cannot adjust.

d) Foundations of high rise buildings are kept wide because more is the area of contact, less is the pressure efforts. So, foundations are wide so as to decrease the possibility of the building from falling down.

e) Deep sea divers or high altitude fliers wear special suits so as prevent their body from being crushed by the water pressure. Since water pressure is maximum at deep seas and oceans, therefore, more is the risk of being injured.

f) Walls of a dam are thickened near the base so that the dam can handle the kinetic energy pressure and prevent itself from breaking down, which if not, can lead to flooding.

HOPE IT HELPS...

5 0

3 years ago

As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte

Klio2033 [76]

Answer:

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

Em_f = U = q V

where the potential is

V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

Em₀ = Em_f

½ m v² = e ( $k \frac{Ze}{r^2}$ )

$r^2 = \frac{ 2 k \ Ze^2}{ 2m}$

with this expression we can find the closest approach distance (r)

3 0

3 years ago