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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?

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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The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

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PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

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Where,

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Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

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Negative symbol is because the force is opposite of the direction of moton.

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3 years ago

A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp

Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by hinge $Fx= \frac{11}{4\sqrt{3} } Mg$

Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

The angle the cable makes with the wall is $\theta = 60^o$

The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

So

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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3 years ago