Physics Questions Set

Study Notes

Physics Questions

Question 25: The expression mv/t represents acceleration.
- Explanation: The equation represents the change in velocity (v) over time (t), which is the definition of acceleration.
Question 26: The speed of the ram on impact is $\sqrt{200}$ m/s.
- Explanation: This question asks about the final velocity of the pile driver right before hitting the post. Applying the conservation of energy, we see that the potential energy at its initial height converts completely into kinetic energy at the impact.
  - Initial potential energy, PE = mgh = 40 kg * 10 m/s² * 5 m = 2000 J.
  - Final kinetic energy, KE = 1/2 * mv² = 2000 J.
  - Solving for velocity, v = sqrt(2KE/m) = sqrt(22000 J/40 kg) = sqrt(200) m/s.
Question 27: The dimensions of the gravitational constant (G) are $M^{-1}L^{3}T^{-2}$.
- Explanation: The gravitational force equation $F = \frac{Gm_1m_2}{r^2}$ helps to find the dimensions of G.
- Force (F) has dimensions of $MLT^{-2}$.
- Mass (m) has dimensions of $M$.
- Distance (r) has dimensions of $L$.
- Combining these, we can solve for the dimensions of G:
- $G = \frac{Fr^2}{m_1m_2} = \frac{MLT^{-2} \cdot L^2}{M \cdot M} = M^{-1}L^{3}T^{-2}$.
Question 28: The dimensions of Power are $ML^{2}T^{-3}$.
- Explanation: Power is defined as the rate of doing work, or how much work is done over a particular time.
- We can find the dimensions of Power by examining the relationship between work and power:
- Power = Work/Time
- Work has dimensions of $ML^2T^{-2}$.
- Time possesses the standard dimension of $T$.
- Combining these, we get the dimensions of Power: $ML^{2}T^{-2}/T = ML^{2}T^{-3}$
Question 29: The dimensions of Energy are $ML^{2}T^{-2}$.
- Explanation: Energy is the capacity to do work.
- Work has dimensions of $ML^2T^{-2}$.
- Therefore, energy has the same dimensions as work, which is $ML^{2}T^{-2}$.
Question 30: The dimensions of the moment of a force about a point are $ML^{2}T^{-2}$.
- Explanation: The moment of a force (torque) is defined as the product of the force and the perpendicular distance from the point to the line of action of the force.
- Force (F) has dimensions of $MLT^{-2}$.
- Distance (r) has dimensions of $L$.
- The moment of force is calculated as the product of force and distance, so its dimensions are $MLT^{-2} \cdot L = ML^{2}T^{-2}$.

Description

Test your understanding of key physics concepts with this interactive quiz. Questions cover topics such as acceleration, velocity, and the gravitational constant, providing explanations to enhance your learning. Perfect for students seeking to reinforce their knowledge in physics.