A 2 kg ball is dropped above the surface of Planet X. If the gravitational field strength at the surface of Planet X is 5 N/kg,
1 answer:
Given data:
* The mass of the ball is 2 kg.
* The gravitational field strength at the surface of planet X is 5 N/kg.
Solution:
The weight of the ball on the planet X is,
where m is the mass of ball, a is the gravitational field strength,
Substituting the known values,
Thus, the weight of the ball on the surface of planet X is 10 N.
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Answer:
KE = 1.75 J
Explanation:
given,
mass of ball, m₁ = 300 g = 0.3 Kg
mass of ball 2, m₂ = 600 g = 0.6 Kg
length of the rod = 40 cm = 0.4 m
Angular speed = 100 rpm=
=10.47\ rad/s
now, finding the position of center of mass of the system
r₁ + r₂ = 0.4 m.....(1)
equating momentum about center of mass
m₁r₁ = m₂ r₂
0.3 x r₁ = 0.6 r₂
r₁ = 2 r₂
Putting value in equation 1
2 r₂ + r₂ = 0.4
r₂ = 0.4/3
r₁ = 0.8/3
now, calculation of rotational energy
KE = 1.75 J
the rotational kinetic energy is equal to 1.75 J
<u>Answer:</u> The Young's modulus for the wire is
<u>Explanation:</u>
Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.
The equation representing Young's Modulus is:
where,
Y = Young's Modulus
F = force exerted by the weight =
m = mass of the ball = 10 kg
g = acceleration due to gravity =
l = length of wire = 2.6 m
A = area of cross section =
r = radius of the wire = (Conversion factor: 1 m = 1000 mm)
= change in length = 1.99 mm =
Putting values in above equation, we get:
Hence, the Young's modulus for the wire is