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AlladinOne [14]
3 years ago
5

A car slows down uniformly from a speed of 21.0m/s to rest in 6.00s. How far did it travel in that time?

Physics
1 answer:
ss7ja [257]3 years ago
4 0

(sorry something wrong w my keyboard so write each line for the explnation!)

63.0 m

Explanation:

Acceleration of car

=

v

−

u

t

=

0 ms

−

1

−

21.0 ms

−

1

6.00 s

=

−

3.50 ms

−

2

S

=

v

2

−

u

2

2a

S

=

(

0 ms

−

1

)

2

−

(

21.0 ms

−

1

)

2

2

×

−

3.50 ms

−

2

S

=

63.0

m

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\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]              (1)

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r: distance to the charge from the point P

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\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}

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