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Harman [31]
3 years ago
13

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

e of the water. It takes a time of 2.30 s for the boat to travel from its highest point to its lowest, a total distance of 0.660 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.50 m apart. How fast are the waves traveling
Physics
1 answer:
Sedbober [7]3 years ago
5 0

Answer:

v = 1.2 m/s

Explanation:

The wavelength of the waves is given as the horizontal distance between the crests:

λ = wavelength = 5.5 m

Now, the time period is given as the time taken by boat to move from the highest point again to the highest point. So it will be equal to twice the time taken by the boat to travel from highest to the lowest point:

T = Time Period = 2(2.3 s) = 4.6 s

Now, the speed of the wave is given as:

v = f\lambda

where,

v= speed of wave = ?

f = frequency of wave = \frac{1}{T} = \frac{1}{4.6\ s} = 0.217\ Hz

Therefore,

v = (0.217\ Hz)(5.5\ m)\\

<u>v = 1.2 m/s</u>

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A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed vcom,0 = 6.0 m/s
ankoles [38]

Answer:

Explanation:

Radius of the ball is R=11cm=0.11m

Initial speed of the ball is v_{com0}=6.0m/s

Initial angular speed of the ball is \omega = 0

Coefficient of kinetic friction between the ball and the lane

is \mu =0.35

Due to the presence of frictional force, ball moves with

decreasing velocity.

(a)

velocity v_{com0} in terms of \omega is

V_{com0} = -R\omega\\\\=-(0.11m)\omega\\\\= (-0.11\omega)m/s

(b)

Ball's linear acceleration is given by

a=-\mu g\\\\=-(0.35) (9.8 m/s^2)\\\\= -3.43m/s^2

(c)

During sliding, ball's angular acceleration is calculated as

\alpha=-\frac{\tau}{I}\\\\-\frac{\mu mgR}{(\frac{2}{5}mR^2)}\\\\-\frac{2}{5}\frac{\mu g}{R}\\\\-\frac{2}{5}\frac{(0.35)(9.8)}{0.11}\\\\-77.95rad/s^2

(d)

The time for which the ball slides is calculated from the

equation of motion is

V_{cm}= V_{cm0} + at\\\\V_{cm} = V_{cm0} + (-\mu g )t\\\\-0.11\omega=6.0m/s -(3.43m/s^2 )t\\\\-(0.11) (\alpha t) =6.0 m/s - (3.43 m/s^2)\\\\- (0.11)(-77.95 rad/s^2)t = 6.0m/s - (3.43 m/s^2 )t\\\\8.5745t + 3.43t= 6.0\\\\12.0045t = 6.0\\\\t= 0.4998s

(e)

Distance traveled by the ball is

X= V_{com,0}+ \frac{1}{2}at^2\\\\= (6.0m/s)(0.4998 s)+ 0.5(-3.43m/s^2) (0.4998 s)^2\\\\=2.57m

(for)

The speed of the ball when smooth rolling begins is

V_{cm} = V_{com, 0}+ at\\\\=6.0 m/s +(-3.43m/s^2 )(0.4998 s)\\\\= 4.29m/s

5 0
3 years ago
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