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Verizon [17]
3 years ago
10

An alien spacecraft flies directly over a football stadium at a speed of 0.50c. if the proper length of the field is 100 yards,

what field length is measured by the crew of the alien spacecraft?
Physics
1 answer:
Ivahew [28]3 years ago
6 0
Go on way and you will get your answer


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I think 9 is function
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3 years ago
Read 2 more answers
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
3 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
Plz help!
Strike441 [17]
72km/hr = 72000m/3600seconds(we convert km/hr to m/s
= 20m/s
Velocity= 20m/s
In 4seconds,distance covered=20x4= 80m
Therefore,an accident occurs because in 4seconds, the car would have moved past 20m
6 0
2 years ago
N discussing engines, the ratio of output work to input work expressed as a percentage is called
djyliett [7]

The ration of output work to input work expressed as a percentage is called <u>Efficiency</u>.

5 0
2 years ago
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