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Harman [31]
2 years ago
5

A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating a

nd drops onto the first flywheel and sticks to it. Both flywheels have radius R. What is the final angular velocity in terms of the initial angular velocity ω0? Treat each flywheel as a disk (I = (1/2) MR2).
Physics
1 answer:
Contact [7]2 years ago
5 0

Answer:

0.83 ω

Explanation:

mass of flywheel, m = M

initial angular velocity of the flywheel, ω = ωo

mass of another flywheel, m' = M/5

radius of both the flywheels = R

let the final angular velocity of the system is ω'

Moment of inertia of the first flywheel , I = 0.5 MR²

Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²

use the conservation of angular momentum as no external torque is applied on the system.

I x ω = ( I + I') x ω'

0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'

0.5 x MR² x ωo = 0.6 MR² x ω'

ω' = 0.83 ω

Thus, the final angular velocity of the system of flywheels is 0.83 ω.

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A man climbs a ladder. Which two quantities can be used to calculate the energy stored of the man at the top of the ladder.
Dvinal [7]

Answer:The answer must be The weight of the man and the vertical distance moved.

Explanation: you calculate it by the force you applied times the distance you moved

8 0
1 year ago
The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time
8_murik_8 [283]

The concept used in this is circuit analysis using the simplification of resistors and capacitors.

Explanation:

The time constant for each of the circuits in figure A, B, C, D and E. Therefore, rank the length of time the bulbs stay lit from longest to shortest by using the value of time constant for each circuit. The rank of the time constant of the circuit is C > A = E > B > DC  > A = E > B > D.

Capacitance is the central concept in electrostatics and constructed devices called capacitors are essential elements of electronic circuits.

If more charge is placed on the conductor the voltage increases proportionately. The ratio of the charge to the voltage is called the capacitance C of the conductor C= q/v.

The resistance increases if you add resistors in series and decreases if you add them in parallel. on the other hand the capacitors increases if it is added parallel and decreases if added in series. hence the circuit longest time constant takes longest time to discharge.

3 0
2 years ago
Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol
klasskru [66]

The current flowing in silicon bar is 2.02 \times 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

     J = (1.6 \times 10^-19) / 0.001 (104 \times 1200 + 1016 \times 500)

     J = 1012480 \times 10^-16 A / m^2.

or

J = 1.01 \times 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 \times 10^-9 \times 0.002 = 2.02 \times 10^-12

So, the current flowing in silicon bar is 2.02 \times 10^-12 A.  

6 0
2 years ago
A circuit is made of 0.4 ohm wire, 150 ohm bulb and a 120ohm rheo stat connected inseries. Determine the total resistance of the
UNO [17]
Answer:
Total resistance of the circuit is 270.4 ohm

Explanation:
We are given that:
resistance of wire = 0.4 ohm
resistance of bulb = 150 ohm
resistance of rheostat = 120 ohm

We are also given that these components are connected in series. This means that the total resistance is summation of all the series components.

Therefore
Total resistance = 0.4 + 150 + 120 = 270.4 ohm

Hope this helps :)
8 0
2 years ago
The flywheel of a steam engine runs with a constant angular velocity of 150 rev/min. When steam is shut off, the friction of the
xz_007 [3.2K]

Answer:

a) -1.14 rev/min²

b) 9900 rev

c) -9.92×10⁻⁴ m/s²

d) 30.8 m/s²

Explanation:

First, convert hours to minutes:

2.2 h × 60 min/h = 132 min

a) Angular acceleration is change in angular velocity over change in time.

α = (ω − ω₀) / t

α = (0 rev/min − 150 rev/min) / 132 min

α = -1.14 rev/min²

b) θ = θ₀ + ω₀ t + ½ αt²

θ = 0 rev + (150 rev/min) (132 min) + ½ (-1.14 rev/min²) (132 min)²

θ = 9900 rev

c) The tangential component of linear acceleration is:

a_t = αr

First,  convert α from rev/min² to rad/s²:

-1.14 rev/min² × (2π rad/rev) × (1 min / 60 s)² = -1.98×10⁻³ rad/s²

Therefore:

a_t = (-1.98×10⁻³ rad/s²) (0.50 m)

a_t = -9.92×10⁻⁴ m/s²

d) The magnitude of the net linear acceleration can be found from the tangential component and the radial component:

a² = (a_t)² + (a_r)²

The radial component is the centripetal acceleration:

a_r = v² / r

a_r = ω² r

First, convert 75 rev/min to rad/s:

75 rev/min × (2π rad/rev) × (1 min / 60 s) = 7.85 rad/s

Find the radial component:

a_r = (7.85 rad/s)² (0.50 m)

a_r = 30.8 m/s²

Now find the net linear acceleration:

a² = (-9.92×10⁻⁴ m/s²² + (30.8 m/s²)²

a = 30.8 m/s²

5 0
2 years ago
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