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son4ous [18]
3 years ago
9

Please help me**** i need some answers now

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0
Output can not be greater than input because the conversion of energy can not be greater than 100%.
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Light hits a flat white wall. Which of the following choices correctly describes the interaction between light and the wall?
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C) Some of the light's energy is absorbed by the wall, and some of the light is reflected in different directions
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4 years ago
1. A man walks round a park, first walking north for 80m, then turning right and walking
drek231 [11]

Answer:

Total distance travelled = 210m

Explanation:

Distance travelled = 80m + 50m + 10m + 70m

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3 years ago
A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
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1. 27.3 m/s

The velocity of the gazelle at any time is given by:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

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t = 6.5 s

Substituting the data, we find the gazelle's top speed:

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2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m

Which is larger than 30 m: this means that the gazelle covers the 30 m during its accelerated motion. Therefore, we can use again the equation:

d=\frac{1}{2}at^2

And substituting d = 30 m, we find the time:

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s

3. 10.6 s

In this case, the  distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

d' = 200-88.7 =111.3 m

Therefore, the time taken to cover this distance is

t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s

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4 years ago
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D because the valence and oxygen has both to be low unless the valence is a big speed the bond would fail to react the normal way
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