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3 years ago

What are 90 percent of the stars in space called?

1 answer:

6 0

Yes there called stars because when a star in space goes nova it shines in the sky like a little twinkle because it far away the sun is also a star too.<span />

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A space probe produces a radio signal pulse. If the pulse reaches Earth 12.3 seconds after it is emitted by the probe, what is t

adoni [48]

Answer:

Distance = 3.69 × 10^9 m

The distance from the probe to Earth is 3.69 × 10^9 m

Explanation:

Distance from the probe to the Earth can be derived using the simple motion formula;

Distance = speed × time .....1

Since a radio signal uses an electromagnetic wave to transfer signal, it has the same speed as the speed of light.

Speed of radio signal = speed of light = 3.0 × 10^8 m/s

time taken to reach the earth = 12.3 seconds

Substituting the values of speed and time into equation 1;

Distance = 3.0 × 10^8 m/s × 12.3 s

Distance = 36.9 × 10^8 m

Distance = 3.69 × 10^9 m

Note: all electromagnetic radiation have the same speed which is equal to 3.0 × 10^8 m/s

5 0

3 years ago

Setting goals that match your lifestyle helps improve chances that they will be reached.

Tasya [4]

Answer:

TRUE!!!!!!

Explanation:

7 0

3 years ago

Read 2 more answers

What is the answer to this definition ?

Paul [167]

Newton's 2nd law of motion is: <em>Force</em> = (mass) x (acceleration) .

Force is the only way to change an object's velocity.

6 0

3 years ago

A butcher has a beam balance and masses 0.5kg and 2kg.How would he measure 1.5kg of meat on the balance at once

LekaFEV [45]

Answer and Explanation:

The butcher will balance the meat and the 0.5 kg mass on one side of the beam balance and 2 kg mass on the other side of the beam balance.
By so doing, when the beam balances then the amount of meat measured will be exactly 1.5 kg as required. Since the 1.5 kg meat and 0.5 kg mass is equivalent to the 2 kg on the other side of the beam balance.

6 0

3 years ago

A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho

NISA [10]

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u = 50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

v = u + at

0 = 25 - 9.81 x t

t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u = 50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

5 0

3 years ago