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Naily [24]
3 years ago
7

At a sports event, the car starts from iosi, in 5.0 siis acceleration is 5.

Physics
1 answer:
In-s [12.5K]3 years ago
8 0

Question:

At a sports event, the car starts from rest. in 5.0 s its acceleration is 5.0 m/s2.  Calculate the distance travelled by car.

Answer:

62.5 metres

Explanation:

Given

u = 0 -- Initial Velocity

a = 5.0m/s^2 --- acceleration

t =5.0s -- time

Required

Calculate the distance

This question will be solved using the following equation of motion

S = ut + \frac{1}{2}at^2

<em>Where S represents the distance</em>

<em></em>

Substitute values for u, t and a

S = 0 * 5.0 + \frac{1}{2} * 5.0 * 5.0^2

S = 0 + \frac{1}{2} * 5.0 * 25.0

S =  \frac{1}{2} * 5.0 * 25.0

S =  \frac{1}{2} * 125.0

S =   62.5m

<em>Hence, the distance travelled is 62.5m</em>

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For situation (a)

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E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) =  4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I  = \sqrt{(7.89e5)^{2}  + (7.89e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) net magnitude =  1.115e6\frac{N}{C} @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

 E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

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I <em>E </em>I  = \sqrt{(7.88e5)^{2}  + (7.88e5)^{2}} =  1.242e6\frac{N}{C}

∅ = tan^{-1}(\frac{7.88e5}{7.88e5} ) = tan^{-1}(1) = 45°

Thus for (a) and (b) the net magnitude =  1.242e6\frac{N}{C} @ 45° above +x axis

Explanation:

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