Question

1. a metal of mass 10 kg falls freely to the surface of a planet with an acceleration of 1.70 m/s². calculate the magnitude of the weight of the metal.

a 11.7 n
b 16.3 n
c 5.9 n
d 17.0 n

1. the energy stored in a capacitor of capacitance 5 μf is 40 j. calculate the voltage across the terminals of the capacitor

a 400 v
b 2000 v
c 4000 v
d 200 v

1. when an elastic material exceeds its elastic limit, the material would

a return to its original size.
b yield at that point.
c break at that point.
d not extend again.

1. two masses, a and b have the same momentum. mass b can have more kinetic energy than a if it

a is moving faster than a
b has the same mass as a.
c has less mass than a.
d is moving at the same speed as a.

1. the process by which heat is transferred from one point to another in a solid material is called

a radiation.
b conduction.
c convection.
d evaporation.

Explanation:

Question 11

Step1: Recall the formula for weight

Weight \( W \) is given by the formula \( W = m \times g \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (or in this case, the acceleration of free fall on the planet).

Step2: Substitute the given values

We are given that \( m = 10 \, \text{kg} \) and \( g = 1.70 \, \text{m/s}^2 \). Substituting these values into the formula:
\( W = 10 \, \text{kg} \times 1.70 \, \text{m/s}^2 \)
\( W = 17.0 \, \text{N} \) Wait, but let me check again. Wait, the options have 17.0 N as option D? Wait, no, wait the options are A:11.7N, B:13.7N, C:5.9N, D:17.0N. Wait, 101.7 is 17, so D? Wait, maybe I miscalculated? Wait 101.7 is 17, so the weight is 17.0 N. So the answer should be D.

Step1: Recall the formula for energy stored in a capacitor

The energy \( E \) stored in a capacitor is given by the formula \( E=\frac{1}{2}CV^{2} \), where \( C \) is the capacitance and \( V \) is the voltage across the capacitor. We need to solve for \( V \).

Step2: Rearrange the formula to solve for \( V \)

Starting with \( E = \frac{1}{2}CV^{2} \), multiply both sides by 2: \( 2E=CV^{2} \). Then divide both sides by \( C \): \( V^{2}=\frac{2E}{C} \). Take the square root of both sides: \( V = \sqrt{\frac{2E}{C}} \)

Step3: Substitute the given values

We are given \( E = 40 \, \text{J} \) and \( C = 5 \, \mu\text{F}=5\times 10^{- 6}\, \text{F} \). Substitute these values into the formula:
\( V=\sqrt{\frac{2\times40}{5\times 10^{-6}}} \)
First, calculate the numerator: \( 2\times40 = 80 \)
Then, the fraction: \( \frac{80}{5\times 10^{-6}}=\frac{80}{5}\times10^{6}=16\times 10^{6}=1.6\times 10^{7} \)
Then, take the square root: \( V=\sqrt{1.6\times 10^{7}} \). Wait, wait, maybe I made a mistake in the capacitance unit. Wait, 5 μF is \( 5\times10^{-6} \) F. Wait, let's recalculate:

\( V=\sqrt{\frac{2\times40}{5\times 10^{-6}}}=\sqrt{\frac{80}{5\times 10^{-6}}}=\sqrt{16\times 10^{6}}=\sqrt{16}\times\sqrt{10^{6}} = 4\times10^{3}=4000 \, \text{V} \). So the voltage is 4000 V, which is option C.

The elastic limit is the maximum stress a material can withstand and still return to its original shape when the stress is removed. If a material exceeds its elastic limit, it will undergo plastic deformation, meaning it will yield (permanently deform) at that point. Option A is incorrect because returning to original size is within the elastic limit. Option C is incorrect as breaking occurs at the breaking point, not just exceeding elastic limit. Option D is incorrect as the material can still extend (plastically) after exceeding elastic limit.

Answer:

D. 17.0 N

Question 12