What's the answer for question 2?

A cylinder of mass 14.0 kg rolls without slipping on a horizontal surface. At a certain instant its center of mass has a speed o

aev [14]

Answer:

a) 567J

b) 283.5J

c)850.5J

Explanation:

The expression for the translational kinetic energy is,

$E_r = \frac{1}{2} mv^2$

Substitute,

14kg for m

9m/s for v

$E_r = \frac{1}{2} (14) (9)^2\\= 567J$

The translational kinetic energy of the center of mass is 567J

(B)

The expression for the rotational kinetic energy is,

$E_R = \frac{1}{2} Iw^2$

The expression for the moment of inertia of the cylinder is,

$I = \frac{1}{2} mr^2$

The expression for angular velocity is,

$w = \frac{v}{r}$

substitute

1/2mr² for I

and vr for w

in equation for rotational kinetic energy as follows:

$E_R = (\frac{1}{2}) (\frac{1}{2} mr^2)(\frac{v}{r} )^2$

$= \frac{mv^2}{4}$

$E_R = \frac{14 \times 9^2 }{4} \\\\= 283.5J$

The rotational kinetic energy of the center of mass is 283.5J

(c)

The expression for the total energy is,

$E = E_r + E_R\\\\$

substitute 567J for E(r) and 283.5J for E(R)

$E = 567J + 283.5\\= 850.5J$

The total energy of the cylinder is 850.5J

6 0

3 years ago

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What forces control the strength of the tides

Colt1911 [192]

Answer:

Gravity is one major force that creates tides. In 1687, Sir Isaac Newton explained that ocean tides result from the gravitational attraction of the sun and moon on the oceans of the earth (Sumich, J.L., 1996).

Explanation:

I hope this helps.

7 0

2 years ago

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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

2 years ago

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A string is vibrating between two posts as shown above. Students are to determine the speed of the wave within this string. They

kow [346]

Answer:

The distance between two posts

Explanation:

As the students have already measured the amount of time necessary for the wave to oscillate up and down that is on the string between two posts,They have to measure the speed of the wave.

ie v=fλ

<em>They have to measure distance between the posts. </em>

Because wavelength is equal to the distance between the posts ie full length of string.

The frequency of the wave is calculated by

f= 1/T

Where is the time period which the students have calculated ie the amount of time taken to oscillate up and down,

Thus the wave speed is calculated using formula

v=λ/T

6 0

2 years ago

Which situation will cause an object moving at constant speed to speed up

Citrus2011 [14]

Answer:

If external is force applied

Explanation:

8 0

2 years ago

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