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Tpy6a [65]
3 years ago
14

An ideal measuring device is one that does not alter the very measurement it is meant to take. Therefore, in comparison with the

resistance being measured, the resistances of an ideal ammeter and an ideal voltmeter must be, respectively: Select the best answer from the choices provided.
Physics
1 answer:
elena55 [62]3 years ago
3 0

Complete question:

An ideal measuring device is one that does not alter the very measurement it is meant to take. Therefore, in comparison with the resistance being measured, the resistances of an ideal ammeter and an ideal voltmeter must be, respectively: Select the best answer from the choices provided.

a) very small; very small

b) very large; very small

c) very small; very large

d) very large; very large

Answer:

c) very small; very large

Explanation:

Ammeters can be said to be a device which measures the flow of electric current through a conductor. An ideal ammeter is said to have zero internal resistance. This is because there will be little or no voltage drop as electric current flows through it.

Therefore the resistance of an ideal ammeter must be very small.

A voltmeter can be said to be a device that measures the difference in potential difference between two points in a given circuit. The internal resistance of a voltmeter is said to be infinite, which means it could be very large. This means no current will flow through the voltmeter and the measured voltage will have little or no error.

Therefore the resistance of an ideal voltmeter must be very large.

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Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
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Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
3 years ago
Which of the following is a contact force?
Nutka1998 [239]
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Explanation:

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