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34kurt
3 years ago
11

A car is traveling in a race.The car went from the initial velocity of 35 to the final velocity of 65 in 5 seconds what is the a

cceleration
Physics
1 answer:
Ann [662]3 years ago
7 0

Answer:

Acceleration = 6m/s²

Explanation:

<u>Given the following data;</u>

Initial velocity = 35m/s

Final velocity = 65m/s

Time = 5 seconds

To find the acceleration;

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

Acceleration (a) = \frac{final \; velocity  -  initial \; velocity}{time}

Where,

  • a is acceleration measured in ms^{-2}
  • v and u is final and initial velocity respectively, measured in ms^{-1}
  • t is time measured in seconds.

Substituting into the equation, we have

a = \frac{65 - 35}{5}

a = \frac{30}{5}

<em>Acceleration = 6m/s²</em>

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Unpolarized light shines pon a photocell for one hour. then, two polarizers, whose transmission axies are offset by 60, are plac
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ball dropped from top of 50m high cliff stone thrown straight up from bottom with speed of 24. they collide. how far above base
Tresset [83]

Answer:

28.73 m from the base of the cliff collide happen

Explanation:

Equation for ball dropped from 50 cliff is reaches distance x in t seconds isi given by

x=ut+\frac{1}{2} at^2\\\\x= 0 \times t+\frac{1}{2} gt^2 =\frac{1}{2} gt^2......(1)

stone is thrown up from the bottom with speed u=24 m/s . it reaches distance y when stone collide with ball.(g is negative here)

y=24t-\frac{1}{2} gt^2.............(2)

we know that total distance traveled by ball and stone is 50 m

x+y=50 m

adding equation 1 and 2, we get time t

x+y=\frac{1}{2} gt^2+24t-\frac{1}{2}gt^2\\50=24t\\t=2.083 s

substitute this time in equation 2, we can get the required distance where they collide

y=24t-\frac{1}{2} gt^2\\y=24\times 2.083-\frac{1}{2}\times  9.8\times 2.083^2\\y=28.73 m

28.73 m from the base of the cliff collide happen

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