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2 years ago

A car speeds up from 13 m/s to 23 m/s in 30 seconds. What is the acceleration of the car?

1 answer:

7 0

Well, v = u + a×t is the equation.

v: final velocity, which is 23 m/s in this equation.
u: initialo velocity = 13 m/s 
a: acceleration = ? 
t: time = 30s

Your equation would be...

23 = 13 + a×30 
a = (23 - 13) / 30 
a = 1 / 3 
a = 0.333 m/s

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An automobile accelerates from zero to 30 m/s in 6.0 s. The wheels have a diameter of 0.40 m. What is the average angular accele

leva [86]

To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.

Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.

The average angular acceleration

$\alpha = \frac{\omega_f - \omega_0}{t}$

Here

$\alpha$ = Angular acceleration

$\omega_{f,i} =$ Initial and final angular velocity

There is not initial angular velocity,then

$\alpha = \frac{\omega_f}{t}$

We know that the relation between the tangential velocity with the angular velocity is given by,

$v = r\omega$

Here,

r = Radius

$\omega$ = Angular velocity,

Rearranging to find the angular velocity

$\omega = \frac{v}{r}}$

$\omega = \frac{30}{0.20} \rightarrow$ Remember that the radius is half te diameter.

Now replacing this expression at the first equation we have,

$\alpha = \frac{30}{0.20*6}$

$\alpha = 25 rad /s^2$

Therefore teh average angular acceleration of each wheel is $25rad/s^2$

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1 year ago

A turntable that is initially at rest is set in motion with a constant angular acceleration α. What is the magnitude of the angu

bekas [8.4K]

Explanation:

If the turntable starts from rest and is set in motion with a constant angular acceleration α. Let $\omega$ is the angular velocity of the turntable. We know that the rate of change of angular velocity is called the angular acceleration of an object. Its formula is given by :

$\alpha =\dfrac{\omega_f-\omega_i}{t}$

$\alpha =\dfrac{\omega-0}{t}$

$\alpha =\dfrac{\omega}{t}$

$t=\dfrac{\omega}{\alpha }$ ............(1)

Using second equation of kinematics as :

$\theta=\omega_i t+\dfrac{1}{2}\alpha t^2$

$\theta=\dfrac{1}{2}\alpha t^2$

Using equation (1) in above equation

$\theta=\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

In one revolution, $\theta=4\pi$ (in 2 revolutions)

$4\pi =\dfrac{1}{2}\times \dfrac{\omega^2}{\alpha }$

$\omega=\sqrt{8\pi \alpha}$

$\omega=2\sqrt{2\pi \alpha}$

Hence, this is the required solution.

6 0

2 years ago

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Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th

Stels [109]

This question deals with the law of conservation of momentum, which basically says that the total momentum in a system must stay the same, provided there are no outside forces. Since you were given the mass and velocity of the two objects you can find the momentum (p=mv) of each and then add them together to find the total momentum of the system before they collide. This total momentum must be the same after they collide. Since you have the mass and velocity of one of the objects after the collision you can find the its momentum after. Subtract this from the the system total and you will have the momentum of the other object after the collision. Now that you know the momentum of the other object you can find its velocity using p=mv and its mass from before.

Be careful with the velocities. They are vectors, so direction matters. Typically moving to the right is positive (+) and moving to the left is negative (-). It is not clear from your question which direction the objects are moving before and after the collision.

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A fisherman is fishing from a bridge and is using a "50.0-N test line." In other words, the line will sustain a maximum force of

umka21 [38]

Answer:(a) 50 N

(b)38.34 N

Explanation:

Given

Maximum tension(T) in line 50 N

(a)If line is moving up with constant velocity i.e. there is no acceleration

This will happen when Tension is equal to weight of Fish

T-mg=0

T=mg

Maximum weight in this case will be 50 N

(b)acceleration of magnitude $2.98 m/s^2$

T-mg=ma

$50=m\left ( g+a\right )$

$50=m\left ( 12.78\right )$

m=3.91

Therefore weight is $3.91\times 9.8=38.34 N$

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