Answer:
K.E₂ = mg(h - 2R)
Explanation:
The diagram of the car at the top of the loop is given below. Considering the initial position of the car and the final position as the top of the loop. We apply law of conservation of energy:
K.E₁ + P.E₁ = K.E₂ + P.E₂
where,
K.E₁ = Initial Kinetic Energy = (1/2)mv² = (1/2)m(0 m/s)² = 0 (car initially at rest)
P.E₁ = Initial Potential Energy = mgh
K.E₂ = Final Kinetic Energy at the top of the loop = ?
P.E₂ = Final Potential Energy = mg(2R) (since, the height at top of loop is 2R)
Therefore,
0 + mgh = K.E₂ + mg(2R)
<u>K.E₂ = mg(h - 2R)</u>
The deeper you go, the more rock must be supported so the more force is required and the pressure goes up.
Using kinematic equation s=ut + 1/2 at^2(u = initial velocity=0, s=120m, t= 6.32s), 120 = 0(t) + 1/2 a(6.32)^2. a = 120x2/(6.32)^2 = 6m/s^2.
Potential energy = mgh
Potential energy = 2 x 9.8 x 1
Potential energy 19.6 J
Answer:
1) The angle of deflection will be less than 45° ( C )
2) The angle of deflection will be greater than 45° but less than 90° ( E )
Explanation:
1) Assuming that the force applied has a direction which is perpendicular to the Earth's magnetic field
∴ Fearth > Fapplied hence the angle of deflection will be < 45°
2) when the Fearth < Fapplied
the angle of deflection will be : > 45° but < 90°