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Rotating helicopter blades; bolt of lightning; radio waves; solar radiation; space flight

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1. A helicopter has blades of length 3.0m, rotating at 2.0rev/s about a central hub. If the vertical component of Earth's magnetic field is 5.0*10^-5T, what is the emf induced between the blade tip and the central hub?

2. A bolt of lightning depicted in a picture passes 200m from a 100-turn coil oriented as shown. If the current in the lighting bolt falls from 6.02*10^6A to zero in 10.5us, what is the average voltage induced in the coil? Assume that the distance to the center of the coil determines the average magnetic field at the coil's position. Treat the lighting bolt as a long, vertical wire. Here is the answer just need to know how to get it. ANSWER=115kV

3. An important news announcement is transmitted by radio waves to people who are 100km away, sitting next to their radios, and by sound waves to people sitting across the newsroom, 3.0m from the newscaster. Who receives the news first? Explain. Take the speed of sound in air to be 343m/s.

4. The intensity of solar radiation at the top of Earth's atmosphere is 1340W/m^3. Assuming that 60% of the incoming solar energy reaches Earth's surface and assuming that you absorb 50% of the incident energy, make an order-of-magnitude estimate of the amount of solar energy you absorb in a 60 min. sunbath.

5. A possible means of space flight is to place a perfectly reflecting aluminized sheet into Earth's orbit and to use the light from the Sun to push this solar sail. Suppose such a sail, of 6.00*10^4m^2 and mass 6000kg, is placed in orbit facing the Sun. a)What force is exerted on the sail? b) What is the sail's acceleration? c) How long does it take for this sail to reach the Moon, 3.84*10^8m away? Ignore all gravitational effects, and assume a solar intensity of 1340W/m^2. Hint[The radiation pressure by a reflected wave is given by 2(average power per unit area)/c.]
Again I have the answer but need to see how its done.
a)ANSWER=0.536N B)ANSWER=8.93*10^-5M/s^2 C)ANSWER=33.9DAYS

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With very detailed explanations and calculations, the problems are solved.

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1.) At any point on the blade at a distance r from the center of rotation:
v = w*r, where w = angular frequency = 2*pi*f
f= 2 rev/sec
Because induced emf: B*v*l
here v is varying from point to point, therefore emf due to each side of blade (i.e l/2 = 1.5 m):
e = integration(0 to l/2)[B*2*pi*f*r*dr]
the induced emf due to both sides will be either from tip to center or center to tip, i.e two emf's will act as parallel batteries of emf e = int(0 to l/2) [B*2*pi*f*r*dr]
=> e = 2*pi*f*B*(l^2/4)*/2= {pi*f*B*l^2}/4 =
=> e = pi* 2*5*10^(-5)*3^2 /4 = 0.71 mV Ans

2.) Sorry I'm not getting clear idea from the picture you have attached.
In the picture ...

Solution provided by:
  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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