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Crank
3 years ago
5

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The

downward component of the earth's magnetic field at this place is
Physics
1 answer:
balandron [24]3 years ago
6 0

This question is incomplete, the complete question is;

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 970 km/hr above East Lansing. The downward component of the earth's magnetic field at this place is 0.7 × 10⁻⁴ T. (DATA: Assume that the wingspan is 60 meters.)

Answer:

the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts

Explanation:

Given that;

Speed S = 970 KM/hr

downward component of the earth's magnetic field B = 0.7 × 10⁻⁴

wingspan 1 = 60min

Velocity V = (970 × 10³) / 3600 = 269.44 m/s

So Average Induced Voltage E = VBI

we substitute

E = 269.44 × (0.7 × 10⁻⁴) × 60

E = 1.132 Volts

Therefore the average induced voltage between the tips of the wings of a Boeing 747 flying is 1.132 Volts

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Speed of the tip of the minute hand=V=0.0244 cm/s

Explanation:

The angular velocity of the minute hand is given by

\omega= \frac{2\pi}{T}

T= time period of the minute hand=60 min=3600 s

so ω= 2 π/3600 rad/s

Now linear velocity v= r ω

r= radius of minute hand=14 cm

so v= 14 (2 π/3600)

V=0.0244 cm/s

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Explanation:

chlorofluorocarbons

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Find the velociity of a car which travels 35 m to the right over a period of 40 seconds
GalinKa [24]

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the velocity of the car is 0.875 m/s

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8 0
3 years ago
The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

20.45 pC

Final capacitance can be calculated as

C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

Final charge= initial charge

q2=q1 (since the battery is disconnected)

q2=q1= 20.45 pC

Final potential difference

V2= q/C2

= 20.45/0.794

= 26V

6 0
3 years ago
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