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3 years ago

If an object triples its velocity, how does this effect its KE?

1 answer:

3 0

If an object triples its velocity then the kinetic energy become 9 times the initial kinetic energy. i hope this helped!

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A mother and her young child want to play on a seesaw at a playground. The child sits on the end of one side of the seesaw. Wher

EleoNora [17]

Answer:middle

Explanation:

Because it will make the seasaw balanced

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3 years ago

Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.

valentina_108 [34]

The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics

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3 years ago

Can we use momentum to see how fast the earth is going?

Kisachek [45]

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

$v=\frac{p}{m}$

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.

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6 0

3 years ago

All you have to do is find the description for a b c and d words and find the meaning 8 9 10 11 and 12

insens350 [35]

A-11 polar easterlies
b-8 winds blowing between the equator and 30° N and south
c-10
d-9

4 0

3 years ago

) Water flows through a horizontal coil heated from the outside by high-temperature flue gases. As it passes through the coil th

Mademuasel [1]

Explanation:

Formula for steady flow energy equation for the flow of fluid is as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

Now, we will substitute 0 for both $z_{1}$ and $z_{2}$ , 0 for w, 334.9 kJ/kg for $h_{1}$ , 2726.5 kJ/kg for $h_{2}$ , 5 m/s for $V_{1}$ and 220 m/s for $V_{2}$ .

Putting the given values into the above formula as follows.

$m[h_{1} + \frac{V^{2}_{1}}{2}] + z_{1}g] + q = m[h_{1} + \frac{V^{2}_{1}}{2} + z_{1}g] + w$

$1 \times [334.9 \times 10^{3} J/kg + \frac{(5 m/s)^{2}}{2} + 0] + q = 1 \times [2726.5 \times 10^{3} + \frac{(220 m/s)^{2}}{2} + 0] + 0$

q = 6597.711 kJ

Thus, we can conclude that heat transferred through the coil per unit mass of water is 6597.711 kJ.

6 0

3 years ago