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2 years ago

Compared to a stationary galaxy, light from a galaxy that is moving away from earth will appear _____.

2 answers:

3 0

"B" When an object moves away from us, the light is shifted to the red end of the spectrum, as its wavelengths get longer.

nikklg [1K]2 years ago

3 0

Answer:

B) Redder

Explanation:

Think of it like this; a car passes by you, as it goes, it's horn gets louder, and then starts to become quieter as it moves further away. Redshift works the same way. When an galaxy starts to move away from Earth, it's light appears to be redder. (If it was moving closer to Earth, it would most likely become bluer.)

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Juan is studying a group of cells under a microscope. His teacher tells him that the cells either came from the leaf of a tomato

RoseWind [281]

Answer:

chloroplasts

Explanation:

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2 years ago

A submarine is 58.8 m from a whale. The sub sends out a sonar ping to locate the whale. The speed of sound underwater is 1520 m/

MissTica

Answer:

0.08

Explanation:

this problem assume that both of whale and submarine are in rest position or in constant linier motion in same direction and same speed.

The sound will travel from Submarine to the whale and back again to submarine. so the time will be like this

t = 2d/v

t = 2*58.8/1520

t = 117.6/1520

t = 0.077368 s

t ≈ 0.08 s (less then 1 s)

6 0

3 years ago

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

5 0

3 years ago

A ball with a mass of 3.1 kg is moving in a uniform circular motion upon a horizontal surface. The ball is attached at the cente

inessss [21]

Answer:

3.46 seconds

Explanation:

Since the ball is moving in circular motion thus centripetal force will be acting there along the rope.

The equation for the centripetal force is as follows -

$F=\frac{mv^2}{r}$

Where, $m$ is the mass of the ball, $v$ is the speed and $r$ is the radius of the circular path which will be equal to the length of the rope.

This centripetal force will be equal to the tension in the string and thus we can write,

$20.4 = \frac{3.1\times v^2}{2}$

and, $v^2 = 13.16$

Thus, $v = 3.63$ m/s.

Now, the total length of circular path = circumference of the circle

Thus, total path length = 2πr = 2 × 3.14 × 2 = 12.56 m

Time taken to complete one revolution = $\frac{\text {Path length} }{\text {Speed}}$ = $\frac{12.56}{3.63}$ = 3.46 seconds.

Thus, the mass will complete one revolution in 3.46 seconds.

4 0

2 years ago

An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

6 0

2 years ago