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4 years ago

The moon revolves around earth at a fairly constant speed.Is the moon accelerating

1 answer:

7 0

<span>If the moon changes direction, it's because it is orbiting around the earth, and the moon keeps a constant speed. And when the moon turns, it accelerates due to velocity. So yes, the moon accelerates. </span>

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Which liquid will evaporate the fastest water salt water pepsi and gatorade?

MArishka [77]

The liquid that will evaporate the quickest would be water.
<em>~The rest are mixtures, and have more ingrediants in them, therefore the answer should be water. (in which it has no other ingrediants in it but water itself)</em>

6 0

3 years ago

[25 points] Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. T

evablogger [386]

6 meters is left because you subtract 12 meters from 6

5 0

3 years ago

A 3 kg object is attached to a 1000 N/m spring. The spring is compressed 0.10 m and then the spring launches the object horizont

Viktor [21]

Answer:

1.826m/s

Explanation:

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6 0

3 years ago

What remains constant during the trajectory of an object

S_A_V [24]

Answer:

projectile

Explanation:

3 0

3 years ago

A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.

IrinaVladis [17]

Answer:

Approximately $722\; \rm m\cdot s^{-1}$ .

Explanation:

The average speed of a vehicle is calculated as:

$\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}$ .

In this question, the total distance is $2 \times 560\; \rm km = 1120\; \rm km$ .

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

$560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m$ .

$1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m$ .

Time required for the first part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{602\; \rm m\cdot s^{-1}} \approx 930\; \rm s$ .

Time required for the second part of this trip:

$\displaystyle \frac{5.60 \times 10^{5}\; \rm m}{903\; \rm m\cdot s^{-1}} \approx 620\; \rm s$ .

The time required for the entire trip would be approximately $930 + 620 = 1550\; \rm s$ .

Calculate the average speed of this plane:

$\begin{aligned} \text{average speed} &= \frac{\text{total distance}}{\text{total time}} \\ &\approx \frac{1.12\times 10^{6}\; \rm m}{1550\; \rm s} \approx 722\; \rm m \cdot s^{-1}\end{aligned}$ .

6 0

3 years ago