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Alchen [17]
3 years ago
7

Four copper wires of equal length are connected in series. Their cross-sectional areas are 1.6 cm2 , 1.2 cm2 , 4.4 cm2 , and 7 c

m2 . If a voltage of 140 V is applied to the arrangement, determine the voltage across the 1.2 cm2 wire. Answer in units of V.
Physics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

63.8 V

Explanation:

We are given that

A_1=1.6 cm^2=1.6\times 10^{-4} m^2

1 cm^2=10^{-4} m^2

A_2=1.2 cm^2=1.2\times 10^{-4} m^2

A_3=4.4 cm^2=4.4\times 10^{-4} m^2

A_4=7 cm^2=7\times 10^{-4} m^2

Potential difference,V=140 V

We know that

R=\frac{\rho l}{A}

According to question

l_1=l_2=l_3=l_4=l

In series

R=R_1+R_2+R_3+R_4

R=\rho l(\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4})

R=\rho l(\frac{1}{1.6\times 10^{-4}}+\frac{1}{1.2\times 10^{-4}}+\frac{1}{4.4\times 10^{-4}}+\frac{1}{7\times 10^{-4}})

R=\rho l(18284.6)

I=\frac{V}{R}=\frac{140}{\rho l\times 18284.6}

Potential across 1.2 square cm=V_1=IR_1=\frac{140}{\rho l\times 18284.6}\times \rho l(\frac{1}{1.2\times 10^{-4}}=63.8 V

Hence, the voltage across the 1.2 square cm wire=63.8 V

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How much energy becomes unavailable for work in an isothermal process at 440 k, if the entropy increase is 25 j/k?
just olya [345]
Answer: 11,000 J

Explanation:

In an isothermal process,

\text{entropy increase} =  \frac{\text{amount of energy in heat transfer}}{\text{temperature} }  (1)

Note that, the energy used in heat transfer is not available for work. So, the amount of energy unavailable for work is equal to the energy used in heat transfer.

To obtain the amount of energy in heat transfer, we multiply both sides of equation (1) by the denominator of the right side of (1) so that 

amount of energy in heat transfer = (entropy increase)(temperature)
                                                      = (25 J/K)(440 K)
                                                      = 11,000 J

Since the amount of energy unavailable for work is equal to the amount of energy in the heat transfer, therefore the amount of energy unavailable for work is 11,000 J.
8 0
3 years ago
An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi
denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0
3 years ago
Which of the following actions would decrease the energy stored in a parallel plate capacitor when a constant potential differen
Debora [2.8K]

Answer:

increasing the separation between the plates

Explanation:

The increase in the vacuum/separation between the plates in a parallel plate capacitor connected to a constant potential difference decreases the energy stored in the capacitor. the increase in the separation of the plates of a parallel plate capacitor reduces the capacitance of the capacitor because

Q(charge) = CV  V = VOLTAGE , c = capacitance  

E = 1/2 eAV^2/ D  ( ENERGY STORED )

where D = distance between plates, e = dielectric, A = area of capacitor , V = potential difference

6 0
3 years ago
The standard wave format for any wave is wave. When depicting wave in standard wave format, the direction of motion must be rota
german

Answer:

Transverse wave  and Longitudinal wave  and Electromagnetic wave

Explanation:

  • An inverted wave is a wave in which the vibrations of the particles are perpendicular to the direction of wave motion.
  • Longitudinal waves, on the other hand, are waves in which the vibrations of the particles are parallel to the direction of wave motion.
  • Electromagnetic waves are waves that do not require medium media for transmission, including radio waves, microwaves, UV lights, etc.
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4 0
3 years ago
A. Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with
zzz [600]

(a) 10 GHz is the frequency of microwave radiation.

(b) 0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

(a). 1 MHz is the frequency of microwave radiation.

(b)  0.167 ms is required by the microwave to travel between two mountains.

Answer:

Explanation:

a. Frequency is the measure of number of times a same thing will be repeated in a given time interval for a given time. And wavelength is the measure of distance between two successive crests or troughs. So wavelength and frequency are inversely proportional to each other. And velocity of light is the proportionality constant.

So frequency of microwave radiation = Speed of light/Wavelength of radiation

Frequency = \frac{3*10^{8} }{3*10^{-2} }

Frequency = 10^{8+2} = 10^{10}=10 GHz

So 10 GHz is the frequency of microwave radiation.

b). As microwave is a part of light waves, so it will be experiencing the speed of light.

As the speed is 3*10^{8} m/s and the distance between the two mountains is given as 50 km, then time can be calculated as

Time = Distance/Velocity

Time = \frac{50*10^{3} m}{3*10^{8} }=16.67*10^{3-8}=16.67*10^{-5}

So time = 0.167 ms.

Thus, 0.167 ms is required by the microwave to travel between two mountains.

6 0
3 years ago
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