Explanation:
It is known that relation between pressure and density is as follows.
P =
where, P = pressure
= density
g = acceleration due to gravity
h = height
Putting the given values into the above formula as follows.
P =
= 
= 110495000 Pa
Now, relation between pressure and force is as follows.
P =
or, F = PA
F =
= 
Thus, we can conclude that a force of
can be experienced at such depth.
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
Answer:
The time is 
The speed is 
Explanation:
From the question we are told that
The height of the cliff is 
Generally from kinematic equation we have that

before the jump the persons initial velocity is u = 0 m/s
So

=> 
Generally from kinematic equation

=> 
=> 
Answer:
1.995 m
Explanation:
Distance of penny as seen by the person = 5 m
Height of person from water surface = 3.50 m
Apparent depth of penny = 5 - 3.50 = 1.5 m
refractive index of water, n = 1.33
real depth / apparent depth = n
real depth = 1.33 x 1.5 = 1.995 m
Thus, the actual depth of water at that point is 1.995 m.