An electric wall clock has a second hand 15 cm long. at the tip of his hand, what is the magnitude of the velocity?

Physics

1 answer:

Mila [183]2 years ago

7 0

The velocity of the tip of the second hand is 0.0158 m/s

Explanation:

First of all, we need to calculate the angular velocity of the second hand.

We know that the second hand completes one full circle in

T = 60 seconds

Therefore, its angular velocity is:

$\omega = \frac{2\pi}{T}=\frac{2\pi}{(60)}=0.105 rad/s$

Now we can calculate the velocity of a point on the tip of the hand by using the formula

$v=\omega r$

where

$\omega=0.105 rad/s$ is the angular velocity

r = 15 cm = 0.15 m is the radius of the circle (the distance of the point from the centre of rotation)

Substituting,

$v=(0.105)(0.15)=0.0158 m/s$

Learn more about angular motion here:

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A fighter bomber is making a bombing run flying horizontally at 500 knots (256m/s) at an altitude of 100.0m.

Jobisdone [24]

Answer:

(a) t = 4.52 sec

(b) X = 1,156.49 m

Explanation:

Horizontal Launching

If an object is launched horizontally, its initial speed is zero in the y-coordinate and the horizontal component of the velocity $v_o$ remains the same in time. The distance x is computed as .

$\displaystyle x=v_o.t$

(a)

The vertical component of the velocity $v_y$ starts from zero and gradually starts to increase due to the acceleration of gravity as follows

$v_y=gt$

This means the vertical height is computed by

$\displaystyle h=h_o-\frac{gt^2}{2}$

Where $h_o$ is the initial height. Our fighter bomber is 100 m high, so we can compute the time the bomb needs to reach the ground by solving the above equation for t knowing h=0