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3 years ago

True or false: (a) the induced emf in a circuit is proportional to the magnetic flux through the circuit.

1 answer:

3 0

I know this is not going to help much perhpas, but it is proportional to the *change* of the magnetic flux.

So it should be false. Imagine a constant flux, the emf is zero, it's not proportional.

But there may be some misunderstanding in who wrote the question. So, I would choose FALSE.

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5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

Volgvan

Answer:

When primary coil is exited by sin wave,this will result in sin wave in secondary coil as well.According to law,flux induced in the secondary coil will have same waveform as in the primary coil.

5 0

4 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Which of the following wavelengths will produce standing waves on a string that is 3.5 m long?

denpristay [2]

In a string of length L, the wavelength of the n-th harmonic of the standing wave produced in the string is given by:

$\lambda=\frac{2}{n} L$

The length of the string in this problem is L=3.5 m, therefore the wavelength of the 1st harmonic of the standing wave is:

$\lambda=\frac{2}{1} \cdot 3.5 m=7.0 m$

The wavelength of the 2nd harmonic is:

$\lambda=\frac{2}{2} \cdot 3.5 m=3.5 m$

The wavelength of the 4th harmonic is:

$\lambda=\frac{2}{4} \cdot 3.5 m=1.75 m$

It is not possible to find any integer n such that $\lambda=5 m$ , therefore the correct options are A, B and D.

3 0

3 years ago

Read 2 more answers

A potential difference of 107 mV exists between the inner and outer surfaces of a cell membrane. The inner surface is negative r

sergij07 [2.7K]

Answer:

The workdone is $W = 1.712 *10^{-20 } \ J$

Explanation:

From the question we are told that

The potential difference is $V = 107 mV = 107 *10^{-3} \ V$

Generally the charge on $Na^{+}$ is $Q_{Na^{+}} = 1.60 *10^{-19 } \ C$

Generally the workdone is mathematically represented as

$W = Q_{Na^{+}}V$

=> $W = 1.60 *10^{-19 } * 107 *10^{-3}$

=> $W = 1.712 *10^{-20 } \ J$

8 0

3 years ago

A net force of 79 N causes a mass to accelerate at a rate of 3 m/s2. Determine the mass

zmey [24]

Force= mass x acceleration

79=mass(3)

m=79/3

m=26.3 kg

4 0

3 years ago