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3 years ago

Georgia drops a brick from the top of an 80.0 m building. What is the velocity of the brick when it hits the ground?

Physics

1 answer:

geniusboy [140]3 years ago

4 0

Answer:

the velocity of the brick when it hits the ground is 39.60 m/s

Explanation:

Given;

height through which the ball is dropped, h = 80.0 m

the initial velocity of the brick, u = 0

The final velocity of the brick when it hits the ground is calculated as;

v² = u² + 2gh

v² = 0 + 2 x 9.8 x 80

v² = 1568

v = √1568

v = 39.60 m/s

Therefore, the velocity of the brick when it hits the ground is 39.60 m/s

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b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

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where $\theta$ is the angle of the rod with respect to the vertical.

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where

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Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

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$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

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