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3 years ago

How does Ben and Jerry's total displacement compare to their distance traveled?

Physics

1 answer:

Sladkaya [172]3 years ago

5 0

Answer:comparing the total displacement and tota distance covered by Ben Jerry.

Explanation:

Total distance travelled by both is;

m+n=z

Total displacement of both is;

x+y=z

Therefore;

comparing equation 1 and 2;

x+y=m+n

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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

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Answer:

Explanation:

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Due to the presence of frictional force, ball moves with

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(a)

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(b)

Ball's linear acceleration is given by

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(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)

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A vertical spring (spring constant =160 N/m) is mounted on the floor. A 0.340-kg block is placed on top of the spring and pushed

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The angular frequency in a spring-mass system is given by

$\omega=\sqrt{\frac{k}{m}}$

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k is the spring constant

m is the mass

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k = 160 N/m

m = 0.340 kg

So the angular frequency is

$\omega=\sqrt{\frac{160 N/m}{0.340 kg}}=21.7 rad/s$

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$f=\frac{\omega}{2\pi}=\frac{21.7 rad/s}{2\pi}=3.5 Hz$

(b) 0.021 m

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x = A

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The maximum displacement is given by Hook's law:

$F=kA$