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padilas [110]
3 years ago
15

Suppose that you observe light emitted from a distant star to be at a wavelength of 525 nm. The wavelength of light to an observ

er on the distant star is 950 nm. What is the velocity of the star relative to you (in units of the speed of light c), and is it moving towards or away from you?
Physics
1 answer:
Tomtit [17]3 years ago
7 0

Answer:

The velocity of the star is 0.532 c.

Explanation:

Given that,

Wavelength of observer = 525 nm

Wave length of source = 950 nm

We need to calculate the velocity

If the direction is from observer to star.

From Doppler effect

\lambda_{0}=\sqrt{\dfrac{c+v}{c-v}}\times\lambda_{s}

Put the value into the formula

525=\sqrt{\dfrac{c+v}{c-v}}\times950

\dfrac{c+v}{c-v}=(\dfrac{525}{950})^2

\dfrac{c+v}{c-v}=0.305

c+v=0.305\times(c-v)

v(1+0.305)=c(0.305-1)

v=\dfrac{0.305-1}{1+0.305}c

v=−0.532c

Negative sign shows the star is moving toward the observer.

Hence, The velocity of the star is 0.532 c.

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Check body of the explanation

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Ooook, quick theory rushdown. if you're at a depth of h in a tank of a fluid, the pressure is the sum of the atmosferic pressure (if the tank is open on top) plus a term which is the product of acceleration of gravity - about 10 ms^-^2, the density of whatever you're sinking in, and the depth at which you are. In formula, p(h) = p_0 + \rho g h, and the pressure is the same for every point of the tank at the same depth.

At this point, we can start answering!

1a. The pressure at A is - not counting atmosferic pressure - 1000 * 10 * 1 = 10^4 Pa, while in B is 1000*10*2 = 2*10^4 Pa, so it's half of it.

1b. The two points are at the same depth, so the pressure is the same - they would be even if the two cilinders weren't linked!

1c. Ditto. Same depth? same pressure!

1d. Usual equation, this time density is 800. Pressure is 800*10*2 = 1,6*10^4 Pa: Since the density is 4/5 of water, the pressure is also 4/5 of the one exerted by water

2a. The volume is simply the product, so 4m*3m*2m = 24m^3

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2d. \rho gh again, what a surprise! 800 {kg \over m^3} * 10 {N \over kg}} * 2 m = 1,6* 10^4 {N\over m^2} =1.6*10^4 Pa

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4 0
2 years ago
A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo
guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s (1km\h=\frac{5}{18} m/s)

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)

a = 1.83 m/s^2

Force=mass\times acceleration

Force=710 kg\times 1.83 m/s^2

Force=1,299.3 N

The net force acting on the car 1,299.3 N.

8 0
3 years ago
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