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d1i1m1o1n [39]
3 years ago
15

Which of the following paraphrases Hubble Law?Select one:A. The greater the distance to a galaxy, the greater its redshift. B. T

he more distant a galaxy is, the younger it appears. C. The older the galaxy appears to us, the more luminous it is. D. The faster the galaxy spins, the more massive and luminous it is. E. The greater the distance to a galaxy, the fainter it is.
Physics
1 answer:
olya-2409 [2.1K]3 years ago
4 0

The correct answer is:

~A. The greater the distance to a galaxy, the greater its redshift.

Hope this helps!!!

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a force of magnitude 30 n stretches a spring 0.83 m from equilibrium. what is the value of the spring constant?
hichkok12 [17]

Answer:

36 N/m

Explanation: I got the answer right on the test ight noo cap

5 0
3 years ago
Charges of +5C and −9C are at a distance of 1m from each other. Which diagram represents the force between the two charges?
weqwewe [10]

the answer is CCCCCCCCCCCCCCCCC

8 0
3 years ago
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What is the velocity of a quarter drooped from a towards after 10secounds
ArbitrLikvidat [17]
U=0 
<span>t=10 </span>
<span>a=9.8m/s/s </span>
<span>v is velocity (the tower must be very high to be able to fall for 10 seconds!!!) </span>

<span>you work out the result now</span>
6 0
2 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
The free-fall acceleration on Mars is 3.7 m/s^2. What length of pendulum has a period of 1.0 s on Earth? What length of pendulum
NemiM [27]

Answer:

Explanation:

Given

Free fall acceleration on mars g_{m}=3.7\ m/s^2

Time Period of pendulum on earth T=1\ s

Time period of Pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

for earth

1=2\pi\cdot \sqrt{\frac{L}{9.8}}

L=\frac{9.8}{(2\pi )^2}

L=0.498\ m

(b)For same time period on mars length is given by

L'=\frac{g_m}{(2\pi )^2}

L'=\frac{3.7}{39.48}

L'=0.0936\ m

L'=9.36\ cm                            

3 0
3 years ago
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