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3 years ago

What was the physical principle used by edwin hubble to demonstrate that the "spiral nebulae" are very distant and large?

1 answer:

4 0

<span>The physical principle used by Edwin Hubble to demonstrate that the "spiral nebulae" are very distant and large was the inverse-square law of light. The Cepheid variable stars in the spiral nebulae obey a period-luminosity relation; by measuring the period of a Cepheid star, its average luminosity can be determined, and then, by comparing this with its average apparent brightness, its distance can be calculated.</span>

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When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The play

saw5 [17]

Answer:

2361.6N

Explanation:

Mass of player = 82kg

Velocity = 1.2m/s

Kinetic energy of player:

= 1/2mv²

= 1/2*82*1.2²

= 41x1.44

= 59.04J

Final kinetic energy = 0

Change in kinetic energy

|∆k| = |0-59.04|

= 59.04

Workdone by the feet = fd

d = 0.025

Fd = 59.04

F = 59.04/0.025

= 2361.6N

This is his average force.

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3 years ago

Question 8

FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0

3 years ago

The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this is powers of ten i

DerKrebs [107]

1.4 times 10 in power of 10

8 0

3 years ago

A worker assigned to the restoration of the Washington Monument is checking the condition of the stone at the very top of the mo

Mnenie [13.5K]

Answer:

Your answer is: K.E = 8.3 J

Explanation:

If the height (h) = 169.2 meters (m) and the mass (m) is 0.005 kilograms (kg) the total energy will be kinetic energy which is equal to the potential energy.

K.E = P.E and also P.E equals to mgh

Then you substitute all the parameters into the formula ↓

P.E = 0.005 × 9.81 × 169.2

P.E = 8.2908 J

So your answer is 8.2908 but if you round it is K.E = 8.3

7 0

3 years ago

How much heat h1 is transferred to the skin by 25.0 g of steam onto the skin? the latent heat of vaporization for steam is l=2.2

elena-s [515]

The heat transferred by the steam to the skin is given by
$Q=m L_v$
where
m is the mass of the steam
$L_v$ is the latent heat of vaporization.

In our problem, the mass of the steam is (converting into kg)
$m=25.0 g=0.025 kg$
while the latent heat of vaporization of the steam is
$L_v = 2.256 \cdot 10^6 J/kg$
Substituting into the previous formula, we find the heat transferred to the skin:
$Q=m L_v = (0.025 kg)(2.256 \cdot 10^6 J/kg)=56400 J = 2.56 \cdot 10^4 J$

8 0

4 years ago