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JAMB UTME - Physics - 2014

Pergunta 1 Relatório

PHCN measures its electrical energy in

W

KWh

Wh

J

Detalhes da Resposta

The PHCN, or Power Holding Company of Nigeria, measures its electrical energy in kilowatt-hours (KWh). This is a unit of energy that is commonly used to measure the amount of electricity that is consumed or generated over a period of time. A kilowatt-hour is equivalent to the amount of energy consumed or generated by a one kilowatt power source in one hour. So, by measuring the electrical energy in kilowatt-hours, PHCN can keep track of how much electricity is being used by its customers and bill them accordingly.

Ler a nota de aula em Electric Cells (JAMB)

Electric Cells

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Pergunta 8 Relatório

A car accelerates uniformly from rest at 3ms-2. its velocity after traveling a distance of 24m is

12ms-1

144ms-1

72ms-1

36ms-1

Detalhes da Resposta

V2 = U2 x 2ax
V2 = 0 x 2(3)(24)
V2 = 144
V =

\sqrt{144}

V = 12ms-1

Ler a nota de aula em Motion (JAMB)

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Pergunta 10 Relatório

The focal length of a concave mirror is 2.0cm. If an object is placed 8.0cm from it, the image is at

2.7m

2.0m

2.3m

2.5m

Detalhes da Resposta

\frac{1}{U} + \frac{1}{V} = \frac{1}{f}

$\frac{1}{8} + \frac{1}{V} = \frac{1}{2}$

$V = \frac{8}{3}$

$V = 2.7 c m$

Ler a nota de aula em Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

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Pergunta 13 Relatório

A wheel and an axle is used to raise a load whose weight is 800N when an effort of 250N is applied. If the radii of the wheel and axle are 800mm and 200mm respectively, the efficiency of the machine is

90%

80%

85%

87%

Detalhes da Resposta

VR = 800/200 = 4

E f f i c i e n c y = \frac{M A}{V R} \times \frac{100}{1}

$E f f i c i e n c y = \frac{L}{E} \times \frac{1}{V R} \times \frac{100}{1}$

$E f f i c i e n c y = \frac{800}{250} \times \frac{1}{4} \times \frac{100}{1}$

= 78.74%
= 80%

Ler a nota de aula em Simple Machines (JAMB)

Simple Machines

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Pergunta 34 Relatório

An object is placed 10m from a pinhole camera of length 25cm. Calculate the linear magnification.

2.5x102

2.5x10-2

2.5x10-1

2.5x101

Detalhes da Resposta

Linear Magnification = \frac{Length of Camera}{Distance of the Object}

$Linear Magnification = \frac{25}{1000}$

$Linear Magnification = 2.5 \times 10^{- 2}$

Ler a nota de aula em Electrostatics (JAMB)

Electrostatics

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Pergunta 35 Relatório

A machine whose efficiency is 75% is used to lift a load of 1000N. Calculate the effort put in to the machine if it has a Velocity ratio of 4.

343.32N

233.33N

333.33N

334.33N

Detalhes da Resposta

\frac{M A}{V R} \times \frac{100}{1} = e f f i c i e n c y

$\frac{L}{E} \frac{1}{V R} \times \frac{100}{1} = 75$

$\frac{1000}{E} \frac{1}{4} \times \frac{100}{1} = 75$

$E = \frac{250 \times 100}{75}$

E = 333.3N

Ler a nota de aula em Simple Machines (JAMB)

Simple Machines

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Pergunta 42 Relatório

A force of 500N1 is applied to a steel wire of cross-sectional area 0.2m2, The tensile stress is

2.5x104Nm-2

1.0x102Nm-2

1.0x103Nm-2

2.5x103Nm-2

Detalhes da Resposta

$S t r e s s = \frac{F o r c e}{A r e a}$
$S t r e s s = \frac{500}{0.2} = 2500$
= 2.5 x 103Nm-3

Ler a nota de aula em Elasticity (JAMB)

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