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butalik [34]
3 years ago
7

An observer on the earth sees a spaceship approaching at 0.54c. The ship then launches an exploration vehicle that, according to

the earth observer, approaches at 0.82c. What is the velocity of the exploration vehicle relative to the spaceship? Give your answer as a decimal fraction of c, i.e. a β factor.
Physics
1 answer:
AURORKA [14]3 years ago
6 0

Answer:

Explanation:

Expression for relative velocity

= \frac{v_1+v_2}{1+\frac{v_1v_2}{c^2} }

= (.54 + .82 )c/ 1+ \frac{.54 \times.82}{1}

= 1.36 c / 1.4428

= .94 c

β = .94

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A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction
steposvetlana [31]

Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

6 0
3 years ago
A light wave has a wavelength of 450 nanometers. What is the frequency of this light?
Zolol [24]

Answer:

Frequency, f=6.67\times 10^{14}\ Hz

Explanation:

Wavelength of a light wave is 450 nm. It is required to find the frequency of this light wave. The speed of light is given by c. So,

c=f\lambda

f is the frequency of this light

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{450\times 10^{-9}}\\\\f=6.67\times 10^{14}\ Hz

So, the frequency of this light is 6.67\times 10^{14}\ Hz.

3 0
3 years ago
Write the percent as a fraction or a mixed number in simplest form. <br><br><br> 7%
svetlana [45]
The actual answer is 7/100
4 0
3 years ago
In the 25 ft Space Simulator facility at NASA's Jet Propulsion Laboratory, a bank of overhead arc lamps can produce light of int
Aleonysh [2.5K]

Answer:

a) <em>8.33 x 10^-6 Pa</em>

b) <em>8.23 x 10^-11 atm</em>

c) <em>1.67 x 10^-5 Pa</em>

d) <em>1.65 x 10^-10 atm</em>

<em></em>

Explanation:

Intensity of the light I = 2500 W/m^2

speed of light c<u> </u>= 3 x 10^8 m/s

a) we know that the pressure for for a totally absorbing surface is given as

P_{abs} = I/c = 2500/(3 x 10^8) = <em>8.33 x 10^-6 Pa</em>

b) 1 atm = 101325 Pa

P_{abs} = (8.33 x 10^-6)/101325 = <em>8.23 x 10^-11 atm</em>

c) for a totally reflecting surface

P_{ref} = 2I/c = twice the value for totally absorbing

P_{ref}  = 2 x 8.33 x 10^-6 = <em>1.67 x 10^-5 Pa</em>

d)  1 atm = 101325 Pa

P_{ref} = 2 x 8.23 x 10^-11  = <em>1.65 x 10^-10 atm</em>

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