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3 years ago

Which characteristic surface of the moon contains the same rock as the ocean floor

Physics

1 answer:

Ratling [72]3 years ago

3 0

The answer is Maria or Lunar mare because they are both made of basalt. Hope this helps.

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Suppose our experimenter repeats his experiment on a planet more massive than Earth, where the acceleration due to gravity is g

Ne4ueva [31]

Answer:

Explanation:

- Let acceleration due to gravity @ massive planet be a = 30 m/s^2

- Let acceleration due to gravity @ earth be g = 30 m/s^2

Solution:

- The average time taken for the ball to cover a distance h from chin to ground with acceleration a on massive planet is:

t = v / a

t = v / 30

- The average time taken for the ball to cover a distance h from chin to ground with acceleration g on earth is:

t = v / g

t = v / 9.81

- Hence, we can see the average time taken by the ball on massive planet is less than that on earth to reach back to its initial position. Hence, option C

7 0

3 years ago

Lolz plz help me with this it’s due soon

White raven [17]

Answer:

Explanation:

c slowing down since the distance is going down and the time is moving up.

7 0

3 years ago

A student is asked to determine the work done on a block of wood when the block is pulled horizontally using an attached string.

Lerok [7]

Answer:

D. Graphing the force as a function of distance and calculating the area under the curve.

Explanation:

5 0

3 years ago

Read 2 more answers

What is the difference between 5 mL of water and 5.0 mL of water?

ad-work [718]

There is no difference. 5 is the same as 5.0

6 0

4 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

2 years ago