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Ann [662]
2 years ago
7

"A uniform cylinder of mass M and radius R is rolling without slipping. The velocity of its center of mass is v. What is the cyl

inder's total kinetic energy?"
Physics
1 answer:
Goshia [24]2 years ago
6 0

Answer:

Explanation:

Given

mass of cylinder is M

radius R

velocity of center of mass is v

As there is no slipping therefore cylinder will rotate as well as translate

Moment of inertia of cylinder I=\frac{MR^2}{2}

Kinetic Energy of cylinder K.E.=\frac{1}{2}Mv^2

Rotational energy R.E.=\frac{1}{2}I\omega ^2

for rolling

v=\omega \times r

where \omega =angular\ velocity\ of\ cylinder

R.E.=\frac{1}{2}\times \frac{MR^2}{2}\times (\frac{v}{R})^2=\frac{1}{4}Mv^2

Total kinetic Energy =\frac{1}{2}Mv^2+\frac{1}{4}Mv^2=\frac{3}{4}Mv^2

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It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and
Trava [24]

Answer:

Explanation:

kinetic energy = 14.1 MJ = 14.1 x 10⁶ J

Let radius of flywheel be r .

volume of flywheel = π r² x t where t is thickness

= 3.14 x r² x .113 m³

= .04 r² m³

mass = volume x density

= .04 r² x 7800 = 312.73 r²kg

moment of inertia I = 1 / 2 mass x radius²

= .5 x 312.73 r² x r²

= 156.37 r⁴ kg m²

angular velocity ω = 2π x 93/60

= 9.734 rad /s

kinetic energy = 1/2 Iω² where ω is angular velocity

= .5 x 156.37 r⁴ x 9.734²

= 7408.08 r⁴

Given

7408.08 r⁴ =  14.1 x 10⁶

r⁴ = .19 x 10⁴

r = .66 x 10

= 6.60 m .

Diameter = 13.2 m

b )

centripetal acceleration of a point on its rim = ω² r

= 9.734² x 6.6

= 625.35 m /s²

5 0
3 years ago
A hover craft also known as a air cushion vehicle glides on a cushion of air allowing it to travel with equal is land or water.
dimaraw [331]

Answer: M = 1797.75 kg

Explanation:

given parameters are;

speed V = 26.7 M/S.

momentum P = 4.8×10^4 KGM/S.

What was the mass of the V. A-3?

Momentum P is the product of mass and velocity. That is, P = MV

Substitute V and P into the formula

4.8×10^4 = 26.7 × M

Make M the subject of formula

M = 4.8×10^4/ 26.7

M = 1797.75 kg

Therefore, the mass of the V. A-3 was 1797.75 kg

4 0
2 years ago
What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c
Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

\displaystyle{t=D/v}

and from Hubble's Law:

v =H_0D

Therefore:

\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}

With the given value for the Hubble's constant we have:

H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s

and thus,

t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years

6 0
2 years ago
An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2
Gekata [30.6K]

Force on a moving charge is given by formula

\vec F = q(\vec v \times \vec B)

here we know that this force will be maximum when velocity is perpendicular to magnetic field

\vec F = qvB

here we know that

v = 2.0 \times 10^5 m/s

q = 1.6 \times 10^{-19} C

B = 8.5 \times 10^2 T

now we have

F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)

F = 2.72 \times 10^{-11} N

7 0
3 years ago
A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values
Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or \lambda=20.0cm

c. The period, or T, of a wave is found in the equation

f=\frac{1}{T} were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

50.0=\frac{1}{T} so

T=\frac{1}{50.0} and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

f=\frac{v}{\lambda} and since we already have the frequency and we solved for the wavelength already, filling in:

50.0=\frac{v}{20.0} and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

5 0
2 years ago
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