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Andru [333]
1 year ago
7

PLEASE ASAP HELP, ILL GIVE WHATEVER POINTS I CAN BUT PLEASE HELP ASAP ASAP

Physics
1 answer:
dalvyx [7]1 year ago
5 0

(1) The ball is in the air for <u>1.4 seconds.</u>

(2) The horizontal velocity of the ball as it rolls off the table is<u> 6.32 m/s.</u>

(3) The vertical velocity of the ball right before it hits the ground is <u>13.72 m/s.</u>

(4) The horizontal velocity of the ball right before it hits the ground is<u> 6.32 m/s.</u>

(5) The initial vertical velocity as soon as the ball comes of the cliff is <u>13.72 m/s​.</u>

<h3>What is the time of motion of the ball?</h3>

The time of motion of the ball is calculated by applying the following equation.

t = √(2h/g)

where;

  • h is the height of the cliff
  • g is acceleration due to gravity

t =  √(2h/g)

t = √(2 x 9.63 / 9.8)

t = 1.4 seconds

The horizontal velocity of the ball is calculated as follows;

v = d/t

where;

  • d is the horizontal distance travelled by the ball = 8.85 m

v = 8.85 m / 1.4 s

v = 6.32 m/s

The vertical velocity of the ball before it hits the ground is calculated as;

vf = vi + gt

vf = 0 + 9.8 x 1.4

vf = 13.72 m/s

The horizontal velocity of the ball right before it hits the ground is calculated as;

the initial velocity of a projectile = final horizontal velocity

vxf = vxi = 6.32 m/s

The initial vertical velocity as soon as the ball comes off the cliff = final vertical velocity = 13.72 m/s

Learn more about horizontal velocity here: brainly.com/question/24949996

#SPJ1

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Answer:

Explanation:

Given

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An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel tu
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a) The initial angular speed is 209.3 m/s

b) The angular acceleration is -1.74 rad/s^2

c) The angular speed after 40 s is 139.7 rad/s

d) The wheel makes 1501 revolutions

Explanation:

a)

The initial angular speed of the wheel is

\omega_i = 2000 rpm

which means 2000 revolutions per minute.

We have to convert it into rad/s. Keeping in mind that:

1 rev = 2\pi rad

1 min = 60 s

We find:

\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s

b)

To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.

Using the same procedure used in part a),

\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s

Now we can find the angular acceleration, given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_i = 209.3 rad/s is the initial angular speed

\omega_f = 104.7 rad/s is the final angular speed

t = 60 s is the time interval

Substituting,

\alpha = \frac{104.7-209.3}{60}=-1.74  rad/s^2

c)

To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:

\omega' = \omega_i + \alpha t

where we have

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 40 s, we find

\omega' = 209.3 + (-1.74)(40)=139.7 rad/s

d)

The angular displacement of the wheel in a certain time interval t is given by

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

where

\omega_i = 209.3 rad/s

\alpha = -1.74 rad/s^2

And substituting t = 60 s, we find:

\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad

So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,

\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev

Learn more about circular motion:

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