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Elis [28]
2 years ago
5

Waves going through a material is called

Physics
1 answer:
Damm [24]2 years ago
7 0
Material medium electric waves
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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.

8 0
3 years ago
What is weightlessness ?​
VMariaS [17]

Answer:

The state of being free from the effects of gravity.

7 0
3 years ago
The process by which you group things based on their similarities is known as classifying.
dusya [7]
True. Classifying by similarities is the basis for biological classification.
8 0
3 years ago
the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. the s
Travka [436]
When the metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off, this is an example of resistance, which provides light and heat. 
3 0
3 years ago
How much money would be saved by turning off one 100.0-W lightbulb 3.0 h/day for 365 days if the
Pavlova-9 [17]

Answer:

the money that would be saved is $13.14.

Explanation:

Given;

power consumed by the light bulb, P = 100 W = 0.1 kW

time of running the bulb, t = 3 hours for 365 days = 1,095 hours

cost rate of power consumption, C = $0.12 per kWh

Energy consumed by the light bulb for the given days;

E = Pt

E = 0.1 kW  x 1,095 hr

E = 109.5 kWh

Cost of energy consumed = 109.5 kWh   x   $0.12 / kWh

                                            = $13.14

Therefore, the money that would be saved is $13.14.

3 0
3 years ago
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