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Musya8 [376]
3 years ago
14

Spaceship 1 and spaceship 2 have equal masses of 150kg. Spaceship 1 has the speed of 0 m/s and spaceship 2 has a speed of 6 m/s.

They collide and stick together and what is their speed ?
Physics
1 answer:
Anna11 [10]3 years ago
5 0

Answer:

3 m/s

Explanation:

The computation of speed is shown below:-

We will compute the equation by using the law of conservation of momentum; also, the total momentum prior and following the collision must be conserved. Therefore we can write the equation is the following manner:-

m_1 u_1 + m_2 u_2 = (m_1 +m_2)v

where the symbols represent

150 kg is the mass of spaceship 1  = m1

150 kg is the mass of spaceship 2  = m2

0 m/s is the initial velocity of spaceship 1  = u1

6 m/s is the initial velocity of spaceship 2 = u2

and v is the velocity of the 2 ships so that they can collide and combined together

For v we will get the following equation to reach the speed

v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}

=\frac{0+(150)(6)}{150+150}

=3 m/s

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3 0
3 years ago
Which property of light is a constant in a vacuum?
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Answer:

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¿Cuál es la consecuencia que a velocidad de la luz sea constante?

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5 0
2 years ago
The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the
Mama L [17]

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

<em>P </em>sun = <em>I </em><em>sun-earth </em><em>A </em><em>sun-earth</em>

where A = 2.863 10²³×m², and <em>I </em> is 1360 W/m²

<em>P </em>sun =  2.863 10²³ × 1360

<em>P </em>sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

<em>I </em><em>sun-mars = </em><em>P </em>sun / A sun-mars

where <em>P </em>sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

<em>I </em><em>sun-mars =  </em>3.85×10²⁶W / 6.53 × 10²³m²

<em>I </em><em>sun-mars = </em>589.6 ≈ 590 W/m²

<em>I </em><em>sun-mars = </em>590 W/m²

6 0
3 years ago
What's the mass show the work?
Alja [10]
You do this one just like the other one that I just solved for you.

For this one ...

The density of the object is 2.5 gm/cm³.
We know that every cm³ of it we have contains 2.5 gm of mass.
We have to find out how many cm³ we have.

The question tells us:  We have  2.0 cm³.

Each cm³ of space that the object occupies contains 2.5 gm of mass.

So the 2.0 cm³ that we have contains (2 x 2.5 gm) = 5 gms.
That's the mass of our object.
6 0
3 years ago
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