2023 - JAMB Physics Past Questions and Answers - page 6

51

 

How much work is done against the gravitational force on a 3.0 kg object when it is carried from the ground floor to the roof of a building, a vertical climb of 240 m?

A

7.2 kJ

B

4.6 kJ

C

6.8 kJ

D

8.4 kJ

correct option: a

The work done against the gravitational force is indeed calculated using the formula \(W = mgh\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity (approximately \(9.8 m/s^2\) on Earth), and \(h\) is the height.

Substituting the given values:

\[ W = 3.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 240 \, \text{m} \]

\[ W = 7.2 \, \text{kJ} \]

Therefore, the correct answer is 7.2 kJ

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52

 

Which of the following is a type of incandescent light source?

A

Fluorescent lamp

B

LED lamp

C

Tungsten filament lamp

D

Neon lamp

correct option: c

The type of incandescent light source among the options is Tungsten filament lamp.

Incandescent light sources produce light by heating a filament until it becomes hot and glows. Tungsten filament lamps are a common example of incandescent light sources, where the filament is made of tungsten.

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53

 

The electrolyte used in the Nickel-Iron (NiFe) accumulator is

A

dilute tetraoxosulphate(VI) acid

B

barium chloride solution

C

potassium hydroxide solution

D

sodium hydroxide solution

correct option: c

The electrolyte used in the Nickel-Iron (NiFe) accumulator is Potassium hydroxide solution.

Nickel-Iron accumulators, also known as Edison batteries, use potassium hydroxide (KOH) solution as the electrolyte.

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54

 

A travelling wave of amplitude 0.80 m has a frequency of 16 Hz and a wave speed of 20 \(ms-1\)

Calculate the wave number of the wave.

A

3

B

4

C

5

D

2

correct option: c

Given:
- Amplitude (\(A\)) = 0.8 m
- Frequency (\(f\)) = 16 Hz
- Wave speed (\(v\)) = 20 \(m/s\)

First, calculate the wavelength (\(\lambda\)) using the formula \(v = f\lambda\):

\[ \lambda = \frac{v}{f} = \frac{20 \, \text{m/s}}{16 \, \text{Hz}} = 1.25 \, \text{m} \]

Then, calculate the wave number (\(k\)) using the formula \(k = \frac{2\pi}{\lambda}\):

\[ k = \frac{2 \pi}{1.25} \approx 5 \, \text{m}^{-1} \]

Therefore, the correct answer is 5 (to 1 significant figure)

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