202 Electrical Science (Test Yourself Questions) and "Worked"

Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

a) Volume of the load

• 12m²

• 12m³

• 12 metres squared

• 12 metres cubed

Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

b) Mass of the load (1dp)

Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

c) Weight or force of the load (Nearest whole number)

Answer explanation

Weight or force of the load

force = mass x acceleration

So 15.6x9.81=153N

EXAM TIP

Remember, acceleration here is due to the pull of

gravity and is, on average, 9.81 m/s² on Earth.

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Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

d) Energy to raise the load (Nearest whole number)

Answer explanation

energy = force × distance

energy = 153 × 72 = 11016J

energy=acceleration×mass×distance

energy = 9.81×15.6×72=11018.592J

∴ The answer is 11019 or 11016 depending on where you start rounding your calculations.

Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

e) Power to raise the load (1dp)

Answer explanation

Worked Example

A load, having a density of 1.3 kg/m³ and dimensions of

2 m length, 2 m width and a depth of 3 m, needs to be

lifted a distance of 72 m in a time of 3 minutes. Calculate

the following values:

f) Speed or velocity of the load (1dp)

Answer explanation

1. A load, having a mass of 140 kg, is to be raised a distance of 80 m in 2 minutes. How much power is needed to complete this task? (1dp)

Worked example

Using Ohm's Law, what is the power of a circuit having 45 MV and 50 kA?

So: 45 MV is 45,000,000 V (or 45 x 10⁶ V)

And 50 kA is 50,000 A (or 50 x 10³ A)

So, on a calculator: 4 5 x10⁶ X 50 x10³ =2.25 x10¹² or 2.25 TW

(or even 2,250,000,000,000 W).

• 2,250,000,000,000 W

• 2.25 TW

• 2.25 x10¹² W

• 2250 x10⁹ W

EXAM TIP

If your calculator gives you an answer in a format that doesn’t suit, use the "ENG" button, as this expresses the same answer into different indices. Use the "SHIFT" button directly to move down in indices and shift "ENG" to move up the indices.

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