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Inessa [10]
3 years ago
7

What is the defining characteristic of a water cycle?

Physics
1 answer:
pav-90 [236]3 years ago
7 0
<span>It has the same starting and ending point</span>
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The model of the atom proposed by Greek philosophers appears similar to the model proposed centuries later by Dalton. What was t
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The main difference between the model of the atom proposed by Greek philosophers and the model proposed centuries later by Dalton is that the Greek one was mainly speculative and philosophical - it wasn't based on real evidence, but on their suggestions and thoughts about the matter. On the other hand, Dalton had the means to prove his theory using viable evidence, not just speculations.
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The efficiency of a modern bicycle is 95% if you exert 200 Joules of input work pedaling a bicycle on level ground what is the u
Delvig [45]

Answer:

95 J

Explanation:

You can calculate efficiency by dividing useful output by total input, then multiplying it to 100.

So the foumula goes like:

Efficiency= (Useful output/Total input)x100

In this question,

Efficiency= 95%

Useful output= x

Total input= 200

Therefore;

95=(x/200)x100

0.95=x/100

x=0.95x100

x=95 Joules

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3 years ago
It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration
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Answer:

About 12 seconds

Explanation:

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3 years ago
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diamong [38]

Answer:

tectonic plate movement

Explanation:

6 0
2 years ago
Given the following situation of marble in motion on rolling 10 m/s horizontally from a height of 1.5-m with negligible friction
Norma-Jean [14]

Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

The ball was initially travelling horizontally. In other words, before leaving the table, the vertical velocity of the ball was v_0(\text{vertical}) = 0 \; \rm m \cdot s^{-1}.

The height of the table was 1.5\; \rm m. Therefore, after hitting the floor, the ball would be 1.5\; \rm m \! below where it was before leaving the table. Hence, \Delta h = -1.5\;\rm m.

The equation becomes:

\displaystyle -1.5 = -\frac{9.81}{2} \, t^{2}.

Solve for t:

\displaystyle t = \sqrt{1.5 \times \frac{2}{9.81}} \approx 0.55.

In other words, it would take approximately 0.55\; \rm s for the ball to hit the floor after leaving the table.

Since the air resistance on the ball is negligible, the horizontal velocity of this ball would be constant (at v(\text{horizontal}) =10\; \rm m \cdot s^{-1}) until the ball hits the floor.

The ball was in the air for approximately t = 0.55\; \rm s and would have travelled approximately v(\text{horizontal})\cdot t \approx 5.5\;\rm m horizontally during the flight.

4 0
2 years ago
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