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3 years ago

9

A car changes velocity at a constant acceleration of 2.5m/s to reach 43.7m/s in 2.7 s how fast was the car moving when it began

to accelerate?

1 answer:

Montano1993 [528]3 years ago

8 0

The formula we can use in this case is:

v = v0 + a t

where v is final velocity, v0 is initial velocity, a is acceleration and t is time

So finding for v0:

v0 = v – a t

v0 = 43.7 – (2.5) 2.7

v0 = 36.95 m/s

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V125BC [204]

Answer: choices a and b

Explanation:

Telescope can be defined as am optical instrument which is designed to observe the distant objects clear and nearer. It comprises of arrangement of lenses which allow the rays of light to be collected. The collected light is focused and the image so produced is magnified in the form of an image. The telescopes are prepared and manufactured on mountains top as this will help in preventing the distortion of light obtain from the star due to the fluctuation of air mass in the atmosphere. The atmospheric distortion affects the resolution, and affects the vision. The atmospheric pressure is low at the mountain tops so it will help in better observation of the sky.

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2 years ago

What measures the distance between two consecutive crests of a wave?

Marina86 [1]

Answer A is incorrect
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B is incorrect
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C is incorrect.
the amplitude measures the height of a crest from the middle of the wave to the crest (or trough).

D is the correct answer. That is the distance between 2 crests or 2 troughs

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3 years ago

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
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Power = 1/f = 1/0.275 = +3.6363 Diopeters.
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3 years ago

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

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The maximum amount of work performed is

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Explanation:

The efficiency of a real heat engine is given by the equation:

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where

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$T_H$ is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

$\eta = \frac{W_{max}}{Q_H}$

where

$W_{max}$ is the maximum work done

$Q_H$ is the heat absorbed from the hot reservoir

$Q_H$ can be written as

$Q_H=W_{max}+Q_C$

where

$Q_C$ is the heat released to the cold reservoir

So the previous equation can be also written as

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By combining eq.(1) and (2) we get

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Learn more about work and heat:

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3 years ago

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valina [46]

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3 years ago