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2 years ago

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An electric lamp consumes 60W at 220 volts. How many dry cells of 1.5 V and internal resistance 1 Ohm are required to glow the l

amp?

1 answer:

GenaCL600 [577]2 years ago

3 0

Answer:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

Explanation:

We'll begin by calculating the resistance. This can be obtained as follow:

Power (P) = 60 W

Voltage (V) = 220 V

Resistance (R) =?

P = V²/R

60 = 220² / R

Cross multiply

60 × R = 220²

60 × R = 48400

Divide both side by 60

R = 48400 / 60

R ≈ 807 Ohm

1. Determination of the number of dry cells of 1.5 V required.

Voltage (V) = 220

Dry Cells = 1.5 V

Number of dry cells (n) =?

n = Voltage / Dry cells

n = 60 / 1.5

n = 40

2. Determination of the number of internal resistance of 1 ohm required.

Resistance (R) = 807 Ohm

Internal resistance (r) = 1 ohm

Number of internal resistance (n) =?

n = R/r

n = 807 / 1

n = 807

SUMMARY:

1. Number of dry cells of 1.5 V required is 40.

2. Number of internal resistance of 1 ohm required is 807

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The gravity force between Jupiter and Io will be 6.343 × 10²² N.

<h3>What is Newton's law of gravitation?</h3>

Newton's law of gravity states that each particle having mass in the universe attracts each other particle with a force known as the gravitational force.

Given data;

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The gravitational force is proportional to the product of the masses of the two bodies and inversely proportional to the square of their distance.

The gravitational force is found as;

$\rm F = G \frac{ m_J m_{I_0}}{R^2} \\\\\ F = (6.67\times 10^{-11}) \frac{( (1.9\times 10^{27})\times (8.9\times 10^{22} )} { (421700000)^2}\\\\ F_g = 6.343 \times 10^{22} \ N$

Hence, the gravity force between Jupiter and Io will be 6.343 × 10²² N.

The complete question is

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2 years ago

1) A record is spinning at the rate of 25 rpm. If a ladybug is sitting 10 cm from the

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<h2>distance = 523 cm</h2>

Explanation:

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= 5/12 rev/sec

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Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz

( c ) The tangential speed is v = angular velocity x radius of rotation

The angular velocity ω = 2π x n , where n is the number of rotations per second

Thus angular velocity = 2π x 5/12 = 5π/6 rad/sec

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Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec

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3 years ago

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3 years ago

A 0.500-kg potato is fired at an angle of 80.0° above the horizontal from a PVC pipe used as a "potato gun" and reaches a height

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Answer:

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Also from kinetic equation of motion, velocity and displacement are related using equation

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Where v is final velocity, u is initial velocity, a is acceleration and s is displacement

From the free body diagram attached, final velocity at maximum height is 0 and initial velocity is $usin80^{o}$

Also, the vertical component can be written as

$v_{y}^{2} }=u_{y}^{2} } -2gs$ The negative sign before 2gs means displacement is opposite the gravitational force

Where g is acceleration due to gravity, $u_{y}$ is vertical component of initial velocity and $v_{y}$ is vertical component of final velocity

Since $v_{y}$ is 0

$u_{y}^{2} } =2gs$

s=110 and g is taken as 9.8

$u_{y}=\sqrt{2*9.8*110}=46.43275$

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Also, it's evident that the vertical component of initial velocity is $u_{y}=u_{i}sin \theta$ where $\theta$ is angle of projection and $u_{i}$ is resultant velocity

Making $u_{i}$ the subject we obtain $u_{i}=\frac {u_{y}}{sin \theta}$

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(b)

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$v^{2}=2sa$ and making a the subject we obtain

$a=\frac {v^{2}}{2s}$

$a=\frac {47.15^{2}}{2*0.450}=2470.13ms^{-2}$

Therefore, average acceleration is $2470.13ms^{-2}$

(c)

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Therefore, the average force of potato is

F=0.5*2470.13=1235.06N

F=1235.06N

The weight, W of potato is given by W=mg

Taking R as ratio of average force and weight of potato

$R=\frac {F}{W}=\frac {F}{mg}$ and since F=1235.06, m=0.500kg and g=9.8

$R=\frac {1235.06}{0.500*9.8}$ =252.05

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