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MrRissso [65]
2 years ago
8

A runner begins from rest at the starting line and travles for 6.5 seconds, a runner reaches a speed of 13.4 m/s what is the run

ners acceleration?
Physics
1 answer:
Butoxors [25]2 years ago
3 0

The acceleration of the runner in the given time is 2.06m/s².

Given the data in the question;

Since the runner begins from rest,

  • Initial velocity; u = 0
  • Final velocity; v = 13.4m/s
  • Time elapsed; t = 6.5s

Acceleration of the runner; a = \ ?

<h3>Velocity and Acceleration</h3>

Velocity is the speed at which an object moves in a particular direction.

Acceleration is simply the rate of change of the velocity of a particle or object with respect to time. Now, we can see the relationship from the First Equation of Motion

v = u + at

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

To determine the acceleration of the runner, we substitute our given values into the equation above.

v = u + at\\\\13.4m/s = 0 + (a * 6.5s)\\\\13.4m/s = a * 6.5s\\\\a = \frac{13.4m/s}{6.5s}\\ \\a = 2.06m/s^2

Therefore, the acceleration of the runner in the given time is 2.06m/s².

Learn more about Equations of Motion: brainly.com/question/18486505

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Answer:

a)v=1.77\times 10^6\ m/s

b)v=3.872\times 10^6\ m/s

c)v=5.5\times 10^6\ m/s

d)v=6.7\times 10^6\ m/s

e)v=7.7\times 10^6\ m/s

Explanation:

Given that

d = 2 cm

V = 200 V

u=4\times 10^5\ m/s

We know that

F = E q

F = m a

E = V/d

So

m a =  q .V/d b            

a=\dfrac{q.V}{m.d}                 ---------1

The mass of electron

m=9.1\times 10^{-31}\ kg

The charge on electron

q=1.6\times 10^{-19}\ C

Now by putting the all values in equation 1

a=\dfrac{1.6\times 10^{-19}\times 200}{9.1\times 10^{-31}\times 0.02}\ m/s^2

a=1.5\times 10^{15}\ m/s^2

We know that

v^2=u^2+2as

a)

s = 0.1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.1\times 10^{-2}

v=\sqrt{3.16\times 10^{12}}\ m/s

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b)

s = 0.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 0.5\times 10^{-2}

v=\sqrt{1.5\times 10^{13}}\ m/s

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c)

s = 1 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1\times 10^{-2}

v=\sqrt{3.06\times 10^{13}}\ m/s

v=5.5\times 10^6\ m/s

d)

s = 1.5 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 1.5\times 10^{-2}

v=\sqrt{4.5\times 10^{13}}\ m/s

v=6.7\times 10^6\ m/s

e)

s = 2 cm

v^2=(4\times 10^5)^2+2\times 1.5\times 10^{15}\times 2\times 10^{-2}

v=\sqrt{6.06\times 10^{13}}\ m/s

v=7.7\times 10^6\ m/s

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