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trapecia [35]
3 years ago
10

A kettle is rated at 1 kW, 220 V. Calculate the working resistance of the kettle.

Physics
1 answer:
Anna [14]3 years ago
5 0

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

P=\dfrac{V^2}{R}

R is resistance

R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

E=P\times t

P is power, P=\dfrac{V^2}{R}

E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ

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Schach [20]
Using lens equation;

1/o + 1/i = 1/f; where o = Object distance, i = image distance (normally negative), f = focal length (normally negative)

Substituting;

1/o + 1/-30 = 1/-43 => 1/o = -1/43 + 1/30 = 0.01 => o = 1/0.01 = 99.23 cm

Therefore, the object should be place 99.23 cm from the lens.
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An object's true weight is 123 N. When it is completely submerged in water, its
SashulF [63]

Object true weight is given as

mg = 123 N

now we know that g = 9.8 m/s^2

m* 9.8 = 123

m = \frac{123}{9.8} = 12.55 kg

now when it is complete submerged in water its apparent weight is given as 82 N

apparent weight = weight - buoyancy force

apparent weight = 82 N

weight = 123 N

now we have

82 = 123 - buoyancy force

buoyancy force = 123 - 82 = 41 N

now we also know that buoyancy force is given as

F_b = p_{liq}Vg

41 = 1000*V*9.8

V = \frac{41}{1000*9.8}

V = 4.18 * 10^{-3} m^3

now as we know that mass of the object is 12.55 kg

its volume is 4.18 * 10^-3 m^3

now we know that density will be given as mass per unit volume

density = \frac{m}{V}

density = \frac{12.55}{4.18*10^{-3}

density = 3002.4 kg/m^3

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7 0
3 years ago
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How much heat in kilojoules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.
blagie [28]
Q = m.Lf.
Here, m = 18.6 g
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Substitute their values, 
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We know, 1 KJ = 1000 J
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In short, Your Answer would be Option D

Hope this helps!



3 0
3 years ago
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