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3 years ago

12

Potential difference must be measured in parallel within a circuit, using a device called a voltmeter

1 answer:

grin007 [14]3 years ago

4 0

Answer:

This is true

Explanation:

This is true because in other To measure the potential difference that is across a circuit, we have to place the device called the voltmeter to be in parallel withe the circuit, so as to get a measurement of the difference in energy from one side of the circuit to the other side. We can also refer to the Potential difference as the voltage and it is measured in volts using a voltmeter

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You're carrying a 3.2-m-long, 24kg pole to a construction site when you decide to stop for a rest. You place one end of the pole

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Answer:

Tension of 132N

Explanation:

We need to apply Summatory of Force to find the tension in the hand.

We define te tensión in the hand as $F_2$ and the Tension in fence post as $F_1$ , then

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We apply summatory of moments then

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We can note that,

$1.6 m - 0.35m=1.25m$

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A ray of light incident in water strikes the surface separating water from air making an angle of 10 ° with the normal to the su

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Answer:

a

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b

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c

$\theta_c = 53.05^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta_1 = 10^o$

The refractive index of water is $n_1 = 1.3$

Generally Snell's law is mathematically represented as

$n_1 sin(\theta_1) = n_2 sin(\theta_ 2)$

Here $n_2$ is the refractive index of air with value $n_2 = 1$

$\theta_2$ is the angle of refraction

So

$\theta _2 = sin^{-1}[\frac{n_1 * sin(\theta _1)}{n_2} ]$

=> $\theta _2 = sin^{-1}[\frac{1.3 * sin(10)}{1} ]$

=> $\theta _2 = 13^o$

Given that the angle should not be greater than $\theta _2 =45^o$ then the angle of incidence will be

$\theta _1 = sin^{-1}[\frac{n_2 * sin(\theta _2)}{n_1} ]$

=> $\theta _1 = sin^{-1}[\frac{1 * sin(45)}{1.3} ]$

=> $\theta _1 =32.94^o$

Generally for critical angle is mathematically represented as

$\theta_c = sin^{-1}[\frac{n_2}{n_1} ]$

=> $\theta_c = sin^{-1}[\frac{1}{1.3} ]$

=> $\theta_c = 53.05^o$

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2 years ago