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3 years ago

7

what happens if a voltmeter is connected in series with other components of the circuit (i.e , ammeter, cell, battery, resistor

and wires). Why is it so?

1 answer:

Talja [164]3 years ago

8 0

Answer:the voltmeter measures the potential difference of the circuit,. Voltmeter is a device used to measure potential difference.

Explanation:

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One mol of a perfect, monatomic gas expands reversibly and isothermally at 300 K from a pressure of 10 atm to a pressure of 2 at

Zolol [24]

Answer:

Explanation:

Given

1 mole of perfect, monoatomic gas

initial Temperature $(T_i)=300 K$

$P_i=10 atm$

$P_f=2 atm$

Work done in iso-thermal process $=P_iV_iln\frac{P_i}{P_f}$

$P_i$ =initial pressure

$P_f$ =Final Pressure

$W=10\times 2.463\times ln\frac{10}{2}=39.64 J$

Since it is a iso-thermal process therefore q=w

Therefore q=39.64 J

(b)if the gas expands by the same amount again isotherm-ally and irreversibly

work done is $=P\Delta V$

$V_1=\frac{RT_1}{P_1}=\frac{1\times 0.0821\times 300}{10}=2.463 L$

$V_2=\frac{RT_2}{P_2}=\frac{1\times 0.0821\times 300}{2}=12.315 L$

$\Delta W=1\times (12.315-2.463)=9.852 J$

$\Delta q=\Delta W=9.852 J$

$\Delta U=0$

8 0

3 years ago

Batman chased the joker on his batbike for 20 minutes, traveling at the speed of 30 kilometers per minute. How far did Batman go

AURORKA [14]

Answer:

123 kilometer

Explanation:

none

8 0

2 years ago

You push on a 30 kg box with a force of 120 N. What is the acceleration of the box2

kumpel [21]

Answer:

6

Explanation:

120/30=6

8 0

2 years ago

An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 m/s, what is the frequency of the

bazaltina [42]

Answer:

the frequency of the second harmonic of the pipe is 425 Hz

Explanation:

Given;

length of the open pipe, L = 0.8 m

velocity of sound, v = 340 m/s

The wavelength of the second harmonic is calculated as follows;

L = A ---> N + N--->N + N--->A

where;

L is the length of the pipe in the second harmonic

A represents antinode of the wave

N represents the node of the wave

$L = \frac{\lambda}{4} + \frac{\lambda}{2} + \frac{\lambda}{4} \\\\L = \lambda$

The frequency is calculated as follows;

$F_1 = \frac{V}{\lambda} = \frac{340}{0.8} = 425 \ Hz$

Therefore, the frequency of the second harmonic of the pipe is 425 Hz.

5 0

3 years ago

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0

2 years ago

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