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zimovet [89]
2 years ago
5

A car starts from rest and accelerates uniformly over a time of 7.25 seconds for a distance of 210 m. Determine the acceleration

of the car. What is the variable Vi equal to?
Physics
1 answer:
lara31 [8.8K]2 years ago
8 0

Answer:

Acceleration is 7.990487515m/s²

Initial velocity is 0m.s

Explanation:

s=ut+(1/2)at²

210=0(7.25)+(1/2)a(7.25²)

210=26.28125a

∴a=7.990487515m/s²

'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.

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the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce
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Answer:

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So 24=32+a\times 4

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

6 0
3 years ago
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3 0
3 years ago
The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.
jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

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We need to calculate the electric field value

Using formula of potential difference

\Delta V=Ed

E=\dfrac{\Delta V}{d}

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

E=\dfrac{1.2}{5.0\times10^{-3}}

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Answer:

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