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3 years ago

Question 6 of 10

Physics

2 answers:

user100 [1]3 years ago

8 0

Answer:

the answer is c

Explanation:

dimulka [17.4K]3 years ago

3 0

Answer:

C. It keeps the electric current flowing in one direction.

Explanation:

just got it correct on my quiz.

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A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

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

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

6 0

3 years ago

Which of the following are ohmic materials? a. Batteries Wires Resistors Light bulb filaments b. Batteries, wires, resistors, an

Bingel [31]

Answer:

A., B., and C.

Explanation:

An Ohmic material is a material that obeys Ohm's Law, V = IR.

In contrast, a non-Ohmic material is one that does not obey Ohm's law.

Ohm's law states that the voltage across an electrical object is proportional to the current flowing through it, with the constant of proportionality being Resistance, R (in Ohm's).

The only Non-Ohmic material is the semiconductor, as semiconductors do not obey Ohm's law.

6 0

3 years ago

Find the resistance of an electric light bulb if a current of 0.08 A flows when the potential difference across the bulb is 120

lora16 [44]

U=RI Ohm's law
then R=U/I
=120/0.08
=2250Ω
hope this helps you

8 0

3 years ago

A 2.3 m long wire weighing 0.075 N/m is suspended directly above an infinitely straight wire. The top wire carries a current of

ZanzabumX [31]

Answer:

Explanation:

Magnetic field near current carrying wire

= $\frac{\mu_0}{4\pi} \frac{2i}{r}$

i is current , r is distance from wire

B = 10⁻⁷ x $\frac{2\times49}{r}$

force on second wire per unit length

B I L , I is current in second wire , L is length of wire

= 10⁻⁷ x $\frac{2\times49}{r}$ x 33 x 1

= 3234 x $\frac{10^{-7}}{r}$

This should balance weight of second wire per unit length

3234 x $\frac{10^{-7}}{r}$ = .075

r = $\frac{3234}{.075}$ x 10⁻⁷

= .0043 m

= .43 cm .

5 0

3 years ago

Each inner energy level of an atom has a maximum number of it can hold

CaHeK987 [17]

That would be a maximum of 4 atoms

3 0

3 years ago

Read 2 more answers