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LiRa [457]
4 years ago
8

A test charge is A a very small negative charge with little miee B a point charge of q 100 C C a spbere of charge D. a very amal

l positive charge with little s
Physics
1 answer:
lukranit [14]4 years ago
6 0

Answer:

D. a very small positive charge with little s

Explanation:

A test charge is a very small charge with positive value which do not disturb the electric field exist in the region

So test charge is to find out the strength of electric field that exist in the region.

If the magnitude of test charge is large then it will change the strength of the existing electric field and due to this the value of the force will be altered.

So here in this case the test charge must be small as well as it must be positive nature

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Answer:

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3 years ago
You connect a 100-resistor, a 800-mH inductor, and a 10.0-uF capacitor in series across a 60.0-Hz, 120-V (peak) source. The impe
steposvetlana [31]

Answer:

Impedance, Z = 107 ohms

Explanation:

It is given that,

Resistance, R = 100 ohms

Inductance, L=800\ mH=800\times 10^{-3}\ H=0.8\ H

Capacitance, C=10\ \mu F=10\times 10^{-6}\ F=10^{-5}\ F

Frequency, f = 60 Hz

Voltage, V = 120 V

The impedance of the circuit is given by :

Z=\sqrt{R^2+(X_C-X_L)^2}...........(1)

Where

X_C is the capacitive reactance, X_C=\dfrac{1}{2\pi fC}

X_C=\dfrac{1}{2\pi \times 60\times 10^{-5}}=265.65\ \Omega

X_L is the inductive reactance, X_L={2\pi fL}

X_L={2\pi \times 60\times 0.8}=301.59\ \Omega

So, equation (1) becomes :

Z=\sqrt{(100)^2+(265.65-301.59)^2}

Z = 106.26 ohms

or

Z = 107 ohms

So, the impedance of the circuit is 107 ohms. Hence, this is the required solution.

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If there is an unusually windy month one might anticipate a. Higher tides b.increased wave size c. Unpredictable currents d. Low
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harkovskaia [24]

Answer:

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