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ICE Princess25 [194]

2 years ago

15

Why are distant objects in the universe moving away from Earth? Because the light they emit is redshifted. Because Earth's gravi

ty pushes them away Because the universe expanding Because of a supernova event

1 answer:

GarryVolchara [31]2 years ago

4 0

Distant objects in the universe moving away from Earth because the universe is expanding.

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Please Help! 30 points! ASAP PLZ :)

Savatey [412]

The best logical answer is A

5 0

2 years ago

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A stone is dropped from the

ICE Princess25 [194]

Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

$\\ \sf\longmapsto t^2=100$

$\\ \sf\longmapsto t=10s$

Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

$\\ \sf\longmapsto T=\dfrac{500}{340}$

$\\ \sf\longmapsto T=1.47\approx 1.5s$

Total time:-

$\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s$

8 0

1 year ago

Study the image of the moving car.

gayaneshka [121]

Answer:

While traveling downhill, the car’s potential is <u>increasing</u> and kinetic energy is <u>decreasing</u>

Explanation:

hope this helps!

6 0

1 year ago

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A horse is trotting along pulling a sleigh through the snow. To move the 225 kg sleigh straight ahead at a constant speed, the h

RoseWind [281]

Answer:

a) The net force acting on the sleigh is zero because it is moving at a constant speed.

b) $F_{fr}= \mu mg = F_{pull}$ , then $\mu = \frac{F_{pull}}{mg} = 0.0557$ .

5 0

2 years ago

In order to open the clam it catches, a seagull will drop the clam repeatedly onto a hard surface from high in the air until the

Vedmedyk [2.9K]

Answer:

2.2 s

Explanation:

Hi!

Let's consider the origin of the coordinate system at the ground, and consider that the clam starts with zero velocity, the equation of motion of the clam is given by

$x(t) = 23.1 m - \frac{1}{2}(9.8 m/s^2) t^2$

We are looking for a time t for which x(t) = 0

$0 = 23.1 m - (4.9 m/s^2) t^2$

Solving for t:

$t = \sqrt{\frac{23.1}{4.9}} s = 2.17124 s$

Rounding at the first decimal:

t = 2.2 s

4 0

2 years ago