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2 years ago

6

A star is moving towards the earth with a speed at 90% the speed of light. It emits light, which moves away from the star at the

speed of light. Relative to us on earth, what is the speed of the light moving toward us from the star?

1 answer:

AnnyKZ [126]2 years ago

5 0

Answer:

The speed of light measured in any frame is c = 3.00E8 m/s.

This is one of Einstein's postulates of special relativity.

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A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential h

Sloan [31]

Answer:

77.88 lbm/ft³

Explanation:

Given,

Specific gravity, SG = 1.25

Density of water, ρ = 62.30 lbm/ft³

density of the fluid =

= S.G x ρ_{water}

= 62.30 x 1.25

= 77.88 lbm/ft³

Density of the fluid is equal to 77.88 lbm/ft³

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3 years ago

How are hurricanes and thunderstorms similar? A. Both cover an area as large as 650 km. B. Both involve the formation of large c

Iteru [2.4K]

Both form when warm air rises:)

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3 years ago

Read 2 more answers

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$

$265v_r + 72.8*5.21 = 0$

$v_b = -72.8*5.21/265 = -1.43 m/s$

So the recoil velocity that the raft would have is -1.43 m/s after the swimmer runs off, relative to the shore

7 0

3 years ago

Two stereo speakers mounted 4.52 m apart on a wall emit identical in-phase sound waves. You are standing at the opposite wall of

Delicious77 [7]

The answer is option c) 0.9 m

4 0

3 years ago

If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant temperature is increased from 0.500 kPa to 0.750 kPa, w

uranmaximum [27]

Answer:

200 mL

Explanation:

Given that,

Initial volume, V₁ = 300 mL

Initial pressure, P₁ = 0.5 kPa

Final pressure, P₂ = 0.75 kPa

We need to find the final volume of the sample if pressure is increased at constant temperature. It is based on Boyle's law. Its mathematical form is given by :

$V\propto \dfrac{1}{P}\\\\P_1V_1=P_2V_2$

V₂ is the final volume

$V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{300\times 0.5}{0.75}\\\\V_2=200\ mL$

So, the final volume of the sample is 200 mL.

5 0

3 years ago