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Marianna [84]
3 years ago
13

What is the angular speed of (a) the second hand, (b) the minute hand, and (c) the hour hand of a smoothly running analog watch?

answer in radians per second?
Physics
1 answer:
Agata [3.3K]3 years ago
8 0
A)We know the formula of the angular speed is                 ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is          ω= 2π / (60s)             = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is             T = 1 hr                = 3600sThen the angular speed of the minute hand is         ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is          T = 12hrs             = (12)(3600)sThen the angular speed of the hour hand is         ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
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011 10.0 points
Ulleksa [173]

Answer:

2.47 m

Explanation:

Let's calculate first the time it takes for the ball to cover the horizontal distance that separates the starting point from the crossbar of d = 52 m.

The horizontal velocity of the ball is constant:

v_x = v cos \theta = (25)(cos 35.9^{\circ})=20.3 m/s

and the time taken to cover the horizontal distance d is

t=\frac{d}{v_x}=\frac{52}{20.3}=2.56 s

So this is the time the ball takes to reach the horizontal position of the crossbar.

The vertical position of the ball at time t is given by

y=u_y t - \frac{1}{2}gt^2

where

u_y = v sin \theta =(25)(sin 35.9^{\circ})=14.7 m/s is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

And substituting t = 2.56 s, we find the vertical position of the ball when it is above the crossbar:

y=(14.7)(2.56) - \frac{1}{2}(9.8)(2.56)^2=5.52 m

The height of the crossbar is h = 3.05 m, so the ball passes

h' = 5.52- 3.05 = 2.47 m

above the crossbar.

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The two examples of non contact forces are:

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