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Paha777 [63]
2 years ago
13

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

Physics
1 answer:
HACTEHA [7]2 years ago
5 0

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

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Brut [27]

Answer: 529.9 Hz

Explanation:

Here we need to use the Doppler equation, so we have:

f' = f*(v + v0)/(v - vs)

Here, f is the frequency = 500Hz

v is the velocity of the wave, = 334m/s

v0 is the velocity of the observer = 20m/s

vs is the velocity of the source = 0m/s

Then we have:

f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz

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3 years ago
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Answer:

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Explanation:

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Tresset [83]

Answer

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Katen [24]

Answer:

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Explanation:

From the question we are told that

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The height of the window from the ground is h  =  36.5 \  m

Generally the time taken for the lunch to reach the ground assuming it fell directly under the window is

t  =  \sqrt{\frac{2 *  h }{g} }

=> t  =  \sqrt{\frac{2 *  36.5 }{9,8} }

=> t  =  2.73 \  s

Generally the time taken for the lunch to reach Henrietta is mathematically represented as

T =  t +  t_1

Here t_1 is the time duration that elapsed after Henrietta has passed below the window the value is given as 4 s

Now

T = 2.73  +  4

=> T = 6.73 \  s

Generally the distance covered by Henrietta before catching her lunch is

s=  v  *  T

=> s=  2.70  * 6.73

=> s=  18.17 \  m

Generally the speed with which Bruce threw her lunch is mathematically represented as

v  =  \frac{18.17}{2.73}

v  =  6.66 \  m/s

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