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3 years ago

14

A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one

reaches the bottom first?
A. Small cylinder
B. Large cylinder
C. Both reach the bottom at the same time.

1 answer:

forsale [732]3 years ago

6 0

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

k² / r² = 1/2

acceleration = g sinθ /( 1 + 1/2 )

= 2 g sinθ / 3

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

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Motions need an unbalanced net force to maintain.<br> True or False?

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Answer:

The answer is False

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A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

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Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

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$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

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3 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

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When a potential difference of 10 V is placed across a certain solid cylindrical resistor, the current through it is 2 A. If the

Keith_Richards [23]

Answer:

New Resistance = 0.5556 ohm

Explanation:

Resistance = resistivity * length /area

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