Answer:
K.E = 30,000 J
Explanation:
Given,
The potential energy of the roller coaster car, P.E = 40000 J
The kinetic energy at height h/4, K.E = ?
According to the law of conservation of energy, the total energy of the system is conserved.
At height 'h', the total energy is,
P.E = mgh
K.E = 0
At height 'h/4', the total energy is
P.E + K.E = mgh
P.E = mgh/4
K.E = 1/2 mv²
Therefore,
mgh/4 + 1/2 mv² = mgh
gh/4 + v²/2 = gh
Hence,
v² = 3gh/2
Substituting in the K.E equation
K.E = 1/2 mv²
= 1/2 m (3gh/2)
= 3/4 mgh
= 3/4 x 40000
= 30000 J
Hence, the K.E of the roller coaster car is, K.E = 30000 J
Answer:
The mass of the cart is 150 kg.
Explanation:
Given that,
Mass of a boy, m₁ = 50 kg
Initial speed of boy, u₁ = 10 m/s
Initial speed of car, u₂ = 0 (at rest)
The speed of the cart with the boy on it is 2.50 m/s, V = 2.5 m/s
Let m₂ is the mass of the cart. Using the conservation of momentum as follows :

So, the mass of the cart is 150 kg.
Your question: The strong nuclear force felt by a single proton in a large nucleus _______________________.
Answer: is about the same as that felt by a single proton in a small nucleus.
Answer:
0.84 cm
Explanation:
u = Object distance = 0.35 cm
v = Image distance = -0.6 cm (near point is considered as image distance and negative due to sign convention)
f = Focal length
From lens equation

Focal length of the lens is 0.84 cm