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Gala2k [10]
3 years ago
13

Spaceships A and B are traveling directly toward each other at a speed 0.5c relative to the Earth, and each has a headlight aime

d toward the other ship.
What value do technicians on ship B get by measuring the speed of the light emitted by ship A's headlight?


Answer choices


1).75c


2)1.0c


3)1.5c


4) .5c
Physics
1 answer:
Rainbow [258]3 years ago
5 0
The answer is 2) 1.0c. Light will always propagate through a vacuum at the speed of light “c”; even when moving at a significant fraction of the speed of light, observers will still measure this as the speed of light and the difference is resultant of time dilation.
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1890J

Explanation:

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A 51.0 kg cheetah accelerates from rest to its top speed of 31.7 m/s. HINT (a) How much net work (in J) is required for the chee
andrey2020 [161]

(a) 2.56\cdot 10^4 J

The work-energy theorem states that the work done on the cheetah is equal to its change in kinetic energy:

W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where

m = 51.0 kg is the mass of the cheetah

u = 0 is the initial speed of the cheetah (zero because it starts from rest)

u = 31.7 m/s is the final speed

Substituting, we find

W=\frac{1}{2}(51.0 kg)(31.7 m/s)^2 - \frac{1}{2}(51.0 kg)(0)^2=2.56\cdot 10^4 J

(b) 6.1 cal

The conversion between calories and Joules is

1 cal = 4186 J

Here the energy the cheetah needs is

E=2.56\cdot 10^4 J

Therefore we can set up a simple proportion

1 cal : 4186 J = x : 2.56\cdot 10^4 J

to find the equivalent energy in calories:

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Answer:

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8 0
3 years ago
Calculate the Fermi energy and the conductivity at room temperature for germanium containing 5×10^16 arsenic atoms per cubic cen
Cloud [144]

Answer:

The Fermi energy is 0.568 eV and the conductivity at room temperature is 31.24 (Ωcm)⁻¹

Explanation:

Data given:

Nd = doping concentration = 5x10⁶/cm³

According the mass action law, the hole concentration is:

P_{0} =\frac{n_{i}^{2}  }{n_{0} } ,n_{i}=2x10^{13} /cm^{3}

E_{f} =E_{i} +KTln(\frac{n_{D} }{n_{i} } ),K=8.61x10^{-19} ,KT=0.026eV

E_{i} =\frac{E_{0} }{2} =\frac{0.63}{2} =0.315eV\\E_{f}=0.315+0.203=0.568eV

The conductivity of n-type of semi-conductor is equal to:

The mobility of Germanium is 3900 cm²/Vs

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