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Marta_Voda [28]

3 years ago

15

Why do you think it is so important that all evidence be identified (such as marking the bullets and tagging firearms)? What con

sequences might result if these procedures were not followed?

1 answer:

igor_vitrenko [27]3 years ago

7 0

You could pet the wrong bullets in the wrong gun

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Define the problem. Look around your house or outside. Is there anything that you're wondering about or have questions about?

bekas [8.4K]

Answer:

Water

Explanation:

How do we have water at our disposal, like that's crazy.

6 0

3 years ago

Which items in this picture provide insulation?

Verdich [7]

Answer:wdym insulation if u mean like covering than the ppls heads

Explanation:

5 0

4 years ago

What causes a satellite to move tangentially to a circular path?

yawa3891 [41]

Explanation : We know that two forces acting on the satellite.

The speed of the satellite and the force of the gravity on the satellite then the result , the satellite moves around the earth in a circular motion.

A satellite to move tangentially to a circular path, while revolving around the circular path because the centripetal force is acting on it.

So, we can say gravity is responsible for the motion of the satellite in a circular path.

5 0

3 years ago

Read 2 more answers

Based on the data table,which car won the race?

kobusy [5.1K]

Answer:

car one

Explanation:

3 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

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

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

3 years ago