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Jobisdone [24]
3 years ago
7

A rubber ball of mass 18.5 g is dropped from a height of 1.90 m onto a floor. The velocity of the ball is reversed by the collis

ion with the floor, and the ball rebounds to a height of 1.45 m. What impulse was applied to the ball during the collision?
Physics
1 answer:
kati45 [8]3 years ago
3 0

Answer:

J=0.211kg.m/s

Explanation:

The impulse-momentum theorem states:

J=\Delta p\\J=m(v_a-v-b)\\where:\\m=mass\\v_a=velocity\_after\\v_b=velocity\_before

The velocity before the impact is given by:

(v_b)^2=2.a.\Delta y\\v_b=\sqrt{2*(9.8m/s^2)1.90m}=6.10m/s(-\hat{j})

For the velocity after the impact:

(v_f)^2=(v_a)^2+2.a.\Delta y\\(0)^2=(v_a)^2+2.(-9.8m/s^s).(1.45m)\\\\v_a=\sqrt{2*9.8m/s^2*1.45m}\\v_a=5.33m/s(\hat{j})

so:

J=18.5*10^{-3}kg(5.33m/s(\hat{j})-6.10m/s(-\hat{j}))\\\\J=18.5*10^{-3}kg(5.33m/s(\hat{j})+6.10m/s(\hat{j}))\\\\J=0.211kg.m/s

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Answer:

Balanced forces

Explanation:

Balanced forces are where two forces of equal size act on an object in opposite directions. It means that in each direction, any pushes and pulls are balanced by another force in the opposite direction.

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To avoid an accident, a driver steps on the brakes to stop a 1000kg car traveling at 65km/h. If the braking distance is 35m, how
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To stop the car it would be 100m because if the car is going to 65km/h then it would still be 100km/h
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3 years ago
At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir
Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

\omega = \omega_{o} + \alpha \cdot t

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

Where:

\theta_{o}, \theta - Initial and final angular position, measured in radians.

\omega_{o}, \omega - Initial and final angular speed, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

t - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}

If \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}

\theta-\theta_{o} = 134.88\,rad

The final angular angular speed can be found by the equation:

\omega = \omega_{o} + \alpha \cdot t

If  \omega_{o} = 25\,\frac{rad}{s}, t = 2.40\,s and \alpha = 26\,\frac{rad}{s^{2}}, then:

\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)

\omega = 87.4\,\frac{rad}{s}

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

\Delta \theta = 134.88\,rad + 435\,rad

\Delta \theta = 569.88\,rad

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})

The angular acceleration is now cleared:

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s} and \theta-\theta_{o} = 435\,rad, the angular deceleration is:

\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}

\alpha = -8.780\,\frac{rad}{s^{2}}

Now, the time interval of the Deceleration Phase is obtained from this formula:

\omega = \omega_{o} + \alpha \cdot t

t = \frac{\omega - \omega_{o}}{\alpha}

If \omega_{o} = 87.4\,\frac{rad}{s}, \omega = 0\,\frac{rad}{s}  and \alpha = -8.780\,\frac{rad}{s^{2}}, the time interval is:

t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }

t = 9.954\,s

The total time needed for the grinding wheel before stopping is:

t_{T} = 2.40\,s + 9.954\,s

t_{T} = 12.354\,s

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0
3 years ago
HELPPP!!!! WILL MARK BRAINLIEST
Harlamova29_29 [7]

Answer: if it is in inches than it is 21.844 cm

Explanation:

7 0
2 years ago
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