If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
1 answer:
Answer:
Height, H = 25.04 meters
Explanation:
Initially the ball is at rest, u = 0
Time taken to fall to the ground, t = 2.261 s
Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :
Here, a = g
H = 25.04 meters
So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.
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Passengers in an aircraft are subject to the Normal and Gravity Force acting on them at a low 'orbit', so tiny that it can be many times compared to the same surface of the earth when speaking in general terms.
In a high orbit space vehicle or in the same space, said force decreases considerably or simply disappears, generating the sensation of weightlessness.
Remember that the Force of Gravity is given under the principle
Where,
G = Gravitational Universal constant
M = Mass of the planet
m = mass of the object
r = Distance from center of the planet
When the radius grows considerably the gravitational force begins to decrease.
Answer:
f = 12 cm
Explanation:
<u>Center of Curvature</u>:
The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of mirror.
<u>The Radius of Curvature</u>:
The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of mirror. It is the distance from pole to the center of curvature.
<u>Focal Length</u>:
The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.
The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:
where,
f = focal length = ?
R = Radius of curvature = 24 cm
Therefore,
<u>f = 12 cm</u>