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2 years ago

Arunner starts from rest and accelerates uniformly to a speed of 8.0 meters per

Physics

1 answer:

VladimirAG [237]2 years ago

3 0

Answer:

2.0 m/s/s

Explanation:

The acceleration of an object is the rate of change of velocity of the object.

Mathematically, it is given by:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

Acceleration is a vector, so it has both a magnitude and a direction.

For the runner in this problem, we have:

u = 0 is the initial velocity (he starts from rest)

v = 8.0 m/s is the final velocity

t = 4.0 s is the time taken

Substituting, we find

$a=\frac{8.0-0}{4.0}=2.0 m/s^2$

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The acceleration of a block attached to a spring is given by a=−(0.324m/s2)cos([2.50rad/s]t) a = − ( 0.324 m / s 2 ) c o s ( [ 2

allsm [11]

Answer:

Looks like you have:

a = -.324 cos 2.5 t

In this case ω^2 A = .324

ω = 2.5

f = ω / (2 * pi) = 2.5 / 6.28 = .40 / sec

5 0

2 years ago

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

$W = (44 N)(0.9 m) = 39.6 J$

now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

5 0

2 years ago

Read 2 more answers

A tennis ball connected to a string is spun around in a vertical, circular path at a uniform speed. The ball has a mass m = 0.15

Oksanka [162]

1) 5.5 N

When the ball is at the bottom of the circle, the equation of the forces is the following:

$T-mg = m\frac{v^2}{R}$

where

T is the tension in the string, which points upward

mg is the weight of the string, which points downward, with

m = 0.158 kg being the mass of the ball

g = 9.8 m/s^2 being the acceleration due to gravity

$m \frac{v^2}{R}$ is the centripetal force, which points upward, with

v = 5.22 m/s being the speed of the ball

R = 1.1 m being the radius of the circular trajectory

Substituting numbers and re-arranging the formula, we find T:

$T=mg+m\frac{v^2}{R}=(0.158 kg)(9.8 m/s^2)+(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=5.5 N$

2) 3.9 N

When the ball is at the side of the circle, the only force acting along the centripetal direction is the tension in the string, therefore the equation of the forces becomes:

$T=m\frac{v^2}{R}$

And by substituting the numerical values, we find

$T=(0.158 kg)\frac{(5.22 m/s)^2}{1.1 m}=3.9 N$

3) 2.3 N

When the ball is at the top of the circle, both the tension and the weight of the ball point downward, in the same direction of the centripetal force. Therefore, the equation of the force is

$T+mg=m\frac{v^2}{R}$

And substituting the numerical values and re-arranging it, we find

$T=m\frac{v^2}{R}-mg=(0.158 kg)\frac{5.22 m/s)^2}{1.1 m}-(0.158 kg)(9.8 m/s^2)=2.3 N$

4) 3.3 m/s

The minimum velocity for the ball to keep the circular motion occurs when the centripetal force is equal to the weight of the ball, and the tension in the string is zero; therefore:

$T=0\\mg = m\frac{v^2}{R}$