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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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How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C?
Svet_ta [14]

Answer:

229,098.96 J

Explanation:

mass of water (m) = 456 g = 0.456 kg

initial temperature (T) = 25 degrees

final temperature (t) = - 10 degrees

specific heat of ice = 2090 J/kg

latent heat of fusion =33.5 x 10^(4) J/kg

specific heat of water = 4186 J/kg

for the water to be converted to ice it must undergo three stages:

  • the water must cool from 25 degrees to 0 degrees, and the heat removed would be Q = m x specific heat of water x change in temp

        Q = 0.456 x 4186 x (25 - (-10)) = 66808.56 J

  • the water must freeze at 0 degrees, and the heat removed would be Q = m x specific heat of fusion x change in temp

         Q = 0.456 x 33.5 x 10^(4) = 152760 J

  • the water must cool further to -10 degrees from 0 degrees, and the heat removed would be Q = m x specific heat of ice x change in temp

        Q = 0.456 x 2090 x (0 - (-10)) = 9530.4 J

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3 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
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Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

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        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

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We have the distance to the object p = 24.0 cm, let's calculate

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distance to the negative image is before the lens

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C) the object is after the focal length, therefore, the image is right and virtual

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