Question

1. a metal of mass 10 kg falls freely to the surface of a planet with an acceleration of 1.70 ms⁻². calculate the magnitude of the weight of the metal.

a. 11.7 n
b. 8.3 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 \( m = 10 \, \text{kg} \) and \( g = 1.70 \, \text{m/s}^2 \). Substituting these into the formula:
\( W = 10 \, \text{kg} \times 1.70 \, \text{m/s}^2 \)
\( W = 17.0 \, \text{N} \)

Step1: Recall the formula for energy stored in a capacitor

The energy \( E \) stored in a capacitor is given by \( E=\frac{1}{2}CV^{2} \), 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, calculate the fraction: \( \frac{80}{5\times 10^{-6}}=\frac{80}{5}\times10^{6}=16\times 10^{6} \)
Now, take the square root: \( V=\sqrt{16\times 10^{6}} = 4\times 10^{3}=4000 \, \text{V} \)

The elastic limit of a material is the maximum stress it can withstand while still returning to its original shape when the stress is removed. If a material exceeds its elastic limit, it will yield (undergo plastic deformation) and not return to its original shape. 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 the elastic limit. Option D is incorrect as the material can still extend (plastically) beyond the elastic limit.

Answer:

D. 17.0 N

Question 12