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olga nikolaevna [1]

3 years ago

6

A 49.5 gram super ball traveling at 25.5 m/s bounces off a brick wall and rebounds at 19.5 m/s. a high-speed camera records this

event. if the ball is in contact with the wall for 4.25 ms, what is the magnitude of the average acceleration of the ball during this time interval

1 answer:

OLga [1]3 years ago

8 0

T = 4.25 ms = 4 x 10⁻³ s, the time for rebound
v₁ = 25.5 m/s, the impacting velocty
v₂ = -19.5 m/s, the rebounding velocity (n the opposite directon)

The change in velocity is
v₂ - v₁ = - (25.5+19.5) = -45 m/s

The acceleration is
a = (-45 m/s)/(4 s) = -11.25 m/s²
The negative sign indicates that the final velocity is opposiye to the impact velocty.

Answer: The magnitude of the acceleration is 11.25 m/s²

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Maggie walks to a friends house which is exactly 1500 meters due South. It takes Maggie 45 minutes for the walk and Maggie has t

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Answer:

<em>Explanation below</em>

Explanation:

<u>Speed vs Velocity </u>

These are two similar physical concepts. They only differ in the fact that the velocity is vectorial, i.e. having magnitude and direction, and the speed is scalar, just the magnitude regardless of the direction. They are strongly related to the concepts of displacement and distance, which are the vectorial and scalar versions of the space traveled by a moving object. The velocity can be computed as

$\displaystyle \vec v=\frac{\vec r}{t}$

Where $\vec r$ is the position vector and t is the time. The speed is

$\displaystyle v=\frac{d}{t}$

To compute $\vec r$ , we only need to know the initial and final positions and subtract them. To compute d, we need to add all the distances traveled by the object, regardless of their directions.

Maggie walks to a friend's house, located 1500 meters from her place. The initial position is 0 and the final position is 1500 m. The displacement is

$\vec r=1500\ m \text{ to the south}$

and the velocity is

$\displaystyle \vec v=\frac{1500}{45}=33.33\ m/s\text{ to the south}$

Now, we know Maggie had to make three different turns of direction to finally get there. This means her distance is more than 1500 m. Let's say she walked 500 m in all the turns, then the distance is

$d=1500+500=2000\ m$

If she took the same time to reach her destiny, she would have to run faster, because her average speed is

$\displaystyle v=\frac{2000}{45}=44.44\ m/s$

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3 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

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Explanation:

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Answer:

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