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serious [3.7K]
3 years ago
11

A plane electromagnetic wave, with wavelength 2.8 m, travels in vacuum in the positive direction of an x axis. the electric fiel

d, of amplitude 260 v/m, oscillates parallel to the y axis. what are the (a) frequency, (b) angular frequency, and (c) angular wave number of the wave
Physics
1 answer:
Arte-miy333 [17]3 years ago
3 0
Ok first you gotta know what they are actually even asking you.

Frequency = <em>wavelengths per second
</em>Angular Frequency = <em>radians per second
</em>Wave Number = <em>wavelengths per meter
</em>Angular Wave Number = <em>radians per meter</em>
<em /><em />
Frequency is Easy, just divide the speed of light by the wavelength. This is your answer for A.

Angular frequency is how many radians per second the ELECTRIC FIELD moves through over one second. We know that there are 2π radians in one wavelength.

So to get the angular frequency just multiply the regular frequency by 2π (6.28) to get the frequency in terms of radians per second (<em>instead of waves per second</em><em />). This is your answer for B.

The wave number of a photon is how many waves <em>per meter</em> the photon moves through rather than waves <em>per second</em><em>, </em>similar but different to the 'frequency' calculations above.

Simply find the waves per meter (1 meter divided by the <em>wavelength</em>) and multiply by 2π to get the <em>radians per meter</em><em />. this is your answer for C.

NOTE: make sure your units are all correct, wavelength in meters, frequency in Hz etc...


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2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th
Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = 5.33 × 10^{-5} T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ B = \frac{\mu_{0}i }{2\pi R}

Where B = magnetic field due to long wires, \mu_{0} = 4\pi \times10^{-7}, R = perpendicular distance from wire to given point

From any one wire R_{1}  = 5 cm, R_{2}  = 3 cm

so we write,

∴ B = B_{1} + B_{2}

 B = \frac{\mu_{0} i}{2\pi R_{1} } +  \frac{\mu_{0} i}{2\pi R_{2} }

 B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]

 B = 5.33\times10^{-5}  T

Therefore, the magnitude of magnetic field at given point = 5.33\times10^{-5} T

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Which of the following describes an action-reaction pair?
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Define homeostasis please I need help
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3 years ago
A steam Rankine cycle operates between the pressure limits of 1500 psia in the boiler and 2 psia in the condenser. The turbine i
AlladinOne [14]

Answer:

a. Mass flow rate through the boiler = 5.462lbm/s

b. Power produced by the turbine = 2525.8kW

c. The rate of heat supply in the boiler = 6901.42Btu/s

d. Thermal efficiency of the cycle = 34.3%

Explanation:

In order to provide a solution, we must assume that ;

- The system is operating at a steady condition

- Kinetic and potential energy changes are negligible

Now from steam tables, we calculate specific volume v and enthalpy h as,

h_1 = 95.96Btu/lb (  h_1 = h_f at 2psia )

v_1 = 0.016238ft^3/lb ( v_1 = v_f at 2psia )

w_{p,in} = v_1(P_2-P_1) = 0.016238(1500-2) * \frac{1}{5.404} = 4.501 Btu/lb

w_p = h_2 - h_1\\h_2 = w_p+h_1=4.501+95.96=100.461Btu/lb

h_3 = 1364.0Btu/lb

s_3 = 1.5073Btu/lb.R

( at P_3 = 1500psia & T_3 = 800^0F )

P_4 = 2psia\\S_4 = S_3\\x_4S = \frac{S_4-S_f}{S_{fg}}=\frac{1.5073-0.1783}{1.7374}=0.765

( S_f & S_{fg} when pressure is 2psia)

h_4S = h_f+x_4S*h_{fg}=95.96+(0.765)(1021.0)=877.025Btu/lb

n_T= \frac{h_3-h_4}{h_3-h_4S}\\ h_4=h_3-n_T(h_3-h_4S)=1364.0-0.90(1364.0-877.025)=925.7Btu/lb

Therefore,

q_{in}=h_3-h_2=1364.0-100.461=1263.54Btu/lb\\q_{out}=h_4-h_1=925.7-95.96=829.74Btu/lb\\w_{net}=q_{in}-q_{out}=1263.54-829.74=433.8Btu/lb

To calculate the mass flow rate of steam in the cycle, we use the formula

W_{net}=mw_{net}\\m=\frac{W_{net}}{w_{net}} =\frac{2500}{433.8}=5.763*(\frac{0.94782Btu}{1Kj} )=5.462lb/s

where 1Kj = 0.947817 Btu

The power output and the rate of heat addition are calculated thus,

W_{T,out}=m(h_3-h_4)=(5.462lb/s)*(1364-925.7)Btu/lb*(\frac{1Kj}{0.94782Btu} )\\=5.462*438.3*1.055=2525.8KW

Q_{in}=mq_{in}=5.462(1263.54)=6901.46Btu/s

The thermal efficiency of the cycle can be found thus;

n_{th}=\frac{W_{net}}{Q_{in}} =\frac{2500}{6901.46}*(\frac{0.94782Btu}{1Kj} ) =0.343

= 34.3%

5 0
2 years ago
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