Current is the flow of electrons. The more electrons that flow, then the higher the amperage which measures this flow. If the current is high enough, then the electrical shock may cause damage of some kind.

"Amperage" is abbreviated as "Amp". You may see something like "milliamp" when it comes to measuring this amperage.

7 0

3 years ago

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Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

lora16 [44]

Answer:

Explanation:

Heat energy given out by the soup

= C_v x ( t₂ - t₁ )

= 33 x ( 340 - 300)

= 1320 J

This heat is given to Carnot engine . Efficiency of engine

= (340 - 300 ) / 340

= 40 / 340

2 / 17

This fraction of total heat given is converted into useable work by the engine.

5 0

4 years ago

Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As

Naddik [55]

Answer:

$\lambda = 1.667 nm$

$\theta = 0.8681^o$

Explanation:

From the question we are told that

The distance of separation is $d = 0.220 \ mm = 0.00022 \ m$

The is distance of the screen from the slit is $D = 2.60 \ m$

The distance between the central bright fringe and either of the adjacent bright $y = 1.97 cm = 1.97 *10^{-2}\ m$

Generally the condition for constructive interference is

$d sin \tha(\theta ) = n \lambda$

From the question we are told that small-angle approximation is valid here.

So $sin (\theta ) = \theta$

=> $d \theta = n \lambda$

=> $\theta = \frac{n * \lambda }{d }$

Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as

$y = D * sin (\theta )$

From the question we are told that small-angle approximation is valid here.

$y = D * \theta$

=> $\theta = \frac{ y}{D}$

$\frac{n * \lambda }{d } = \frac{y}{D}$

$\lambda =\frac{d * y }{n * D}$

substituting values

$\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }$

$\lambda = 1.667 *10^{-6}$

$\lambda = 1.667 nm$

In the b part of the question we are considering the next set of bright fringe so n= 2

Hence

$dsin (\theta ) = n \lambda$

$\theta = sin^{-1}[\frac{ n * \lambda }{d} ]$

$\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ]$

$\theta = 0.8681^o$

7 0

4 years ago