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german
3 years ago
10

Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 3.56 kg and rotate with

the same angular speed of 179 rad/s, but they differ in radius. What is the rotational kinetic energy of (a) the smaller cylinder, of radius 0.330 m, and (b) the larger cylinder, of radius 0.714 m?
Physics
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

(a) 3107.98 J

(b) 14530.6 J

Explanation:

mass, m = 3.56 kg

angular speed, ω = 179 rad/s

Moment of inertia of solid cylinder, I = 1/2 mr^2

where, m is the mass and r be the radius of the cylinder.

(a) radius, r = 0.330 m

I = 0.5 x 3.56 x 0.330 x 0.330 = 0.194 kgm^2

The formula for the rotational kinetic energy is given by

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.194 x 179 x 179 = 3107.98 J

(b) radius, r = 0.714 m

I = 0.5 x 3.56 x 0.714 x 0.714 = 0.907 kgm^2

The formula for the rotational kinetic energy is given by

K = \frac{1}{2}I\omega ^{2}

K = 0.5 x 0.907 x 179 x 179 = 14530.6 J

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Answer:

counting tree rings can indicate age

Explanation:

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3 years ago
If the frog lands with a velocity equal to its average velocity and comes to a full stop 0.25s later, what is the frog’s average
Nina [5.8K]

Answer:

Average accelation = -4V

Explanation:

a=\frac{V-V0}{t}

V=0 m/s (because the frog stopped)

V0 = V (average velocity)

t= 0,25 s

So;

a=\frac{V-V0}{t}=\frac{0-V}{0.25}=-4V

4 0
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Sohan said to Geeta I have done my work change the following sentence into indirect​
12345 [234]

Answer:

Sohan told Geeta that i had done my work

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3 years ago
A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image
Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}

=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}

Re-arrange to get i

\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

m=\frac{h'}{h} =-\frac{i}{o}

substitute the values to get the height of image h'

\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm

5 0
3 years ago
A man standing on top of a 30 m tall building throws a brick downwards with a velocity of 12 m/s. Determine the speed of the bri
loris [4]

Answer:

27.1m/s

Explanation:

Given parameters:

Height of the building  = 30m

Initial velocity  = 12m/s

Unknown:

Final velocity  = ?

Solution:

We apply one of the kinematics equation to solve this problem:

         v²  = u²  + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

          v²   = 12²  + (2 x 9.8 x 30)

          v  = 27.1m/s

3 0
3 years ago
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