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Goryan [66]
3 years ago
11

Where is the epicenter of the earthquake located?

Physics
1 answer:
klasskru [66]3 years ago
8 0

Answer:

The Hypocenter

Explanation:

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Suppose you increase your walking speed from 7 m/s to 15 m/s in a period of 3 s. What is your acceleration?
lisov135 [29]

You asked the question twice I answered it on the last one

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3 years ago
As gravity continues to crush the inner core, the fusion of carbon and oxygen begins. What temperature is needed for the fusion
Olin [163]

Answer:

600,000,000 degree C

Explanation:

This stage is the last stage and is refereed to as supernova. In the beginning of this stage, gravity pulls the inner core and crush it, due to which fusion of atoms starts. Carbon and Oxygen fuse together and the temperature is about of 600,000,000 degree C.

The most heavier atom that can be formed out of this fusion is the iron. The moment all the atoms becomes of iron, no further fusion is possible hence that body emits radiation of high intensity and collapse causing a big supernova.

8 0
2 years ago
I will give the Branliest!!! In which body of water can you see your reflected image better, a pond or an ocean? Why? (Please ex
GREYUIT [131]

Answer:

I would say a pond

Explanation:

A pond is more still than an ocean, therefore you could see your reflection better

4 0
2 years ago
Read 2 more answers
From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequ
V125BC [204]

Answer:

55.8 m/s

Explanation:

f = Actual frequency of sound emitted by car

f' = frequency observed when the car moves = (0.86)f

V = Speed of sound = 343 m/s

v = Speed of car

Frequency observed is given as

f' = \frac{Vf}{V+v}

(0.86) = \frac{(343)}{343 + v}\\(0.86) = \frac{(343)}{343 + v}\\294.98 + (0.86) v = 343 \\v = 55.8 ms^{-1}

6 0
2 years ago
A seagull flies at a velocity of 9.00 m/s straight into the wind. (a) if it takes the bird 20.0 min to travel 6.00 km relative t
enot [183]

Here we will the speed of seagull which is v = 9 m/s

this is the speed of seagull when there is no effect of wind on it

now in part a)

if effect of wind is in opposite direction then it travels 6 km in 20 min

so the average speed is given by the ratio of total distance and total time

v_{avg} = \frac{6000}{20*60}

v_{avg} = 5m/s

now since effect of wind is in opposite direction then we can say

V_{net} = v_{bird} - v_{wind}

5 = 9 - v_{wind}

v_{wind}= 4 m/s

Part b)

now if bird travels in the same direction of wind then we will have

v_{net}= v_{bird} + v_{wind}

v_{net} = 9 + 4 = 13 m/s

now we can find the time to go back

time = \frac{distance}{speed}

time = \frac{6000}{13}

time = 7.7 minutes

Part c)

Total time of round trip when wind is present

T = t_1 + t_2

T = 20 + 7.7 = 27.7 min

now when there is no wind total time is given by

T = \frac{6000}{9} + \frac{6000}{9}

T = 22.22 min

So due to wind time will be more

4 0
3 years ago
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