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Law Incorporation [45]
1 year ago
7

A disk of mass m and radius r is initially held at rest just above a larger disk of mass M and radius R that is rotating at angu

lar speed ωi. What is the final angular speed of the disks after the top one is dropped onto the bottom one and they stop slipping on each other? Note: the moment of inertia of a disk of mass M and radius R about an axis through its center and perpendicular to the disk is I = (1/2)MR .
Physics
1 answer:
Leto [7]1 year ago
6 0

Answer:

Explanation:

Moment of inertia of larger disk   I₁ = 1/2 MR²

Moment of inertia of smaller  disk   I₂ = 1/2 m r ²

Initial angular velocity

We shall apply law of conservation of angular momentum .

initial total momentum = final angular momentum

I₁ X ωi  = ( I₁ + I₂ )ωf

1/2 MR² x ωi = 1/2 ( m r² + MR² ) ωf

ωf =  ωi   / ( 1 + m r²/MR² )

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The lighting needs of a storage room are being met by six fluorescent light fixtures, each fixture containing four lamps rated a
Amanda [17]

Answer:

amount of energy  = 4730.4 kWh/yr

amount of money = 520.34 per year

payback period = 0.188 year

Explanation:

given data

light fixtures = 6

lamp = 4

power = 60 W

average use = 3 h a day

price of electricity = $0.11/kWh

to find out

the amount of energy and money that will be saved and simple payback period if the purchase price of the sensor is $32 and it takes 1 h to install it at a cost of $66

solution

we find energy saving by difference in time the light were

ΔE = no of fixture × number of lamp × power of each lamp × Δt

ΔE is amount of energy save and Δt is time difference

so

ΔE = 6 × 4 × 365 ( 12 - 9 )

ΔE = 4730.4 kWh/yr

and

money saving find out by energy saving and unit cost that i s

ΔM = ΔE × Munit

ΔM = 4730.4 × 0.11

ΔM = 520.34 per year

and

payback period is calculate as

payback period = \frac{excess initial cost}{\Delta M}

payback period = \frac{32 + 66}{520.34}

payback period = 0.188 year

8 0
2 years ago
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed?
sweet [91]

Answer:

Speed of the electron will be v=5.896\times 10^7m/sec

Explanation:

We have given that charge on electron e=1.6\times 10^{-19}C

Mass of electron m=9.11\times 10^{-31}kg

Potential difference = V=9.9KV=9.9\times 10^3volt

Now according to energy conservation eV=\frac{1}{2}mv^2

1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2

v=5.896\times 10^7m/sec

4 0
1 year ago
A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance x fro
MArishka [77]

Answer:

Resistance = 3.35*10^{-4} Ω

Explanation:

Since resistance R = ρ\frac{L}{A}

whereas \rho(x) = a + bx^2

resistivity is given for two ends. At the left end resistivity is 2.25* 10^{-8} whereas x at the left end will be 0 as distance is zero. Thus

2.25*10^{-8} = a + b(0)^2\\ 2.25*10^{-8} = a + 0 \\2.25*10^{-8} = a

At the right end x will be equal to the length of the rod, so x = 1.50\\8.50*10^{-8} = (2.25*10^{-8}) + ( b* (1.50)^2 )\\8.50*10^{-8} - (2.25*10^{-8}) = b*2.25\\\frac{6.25*10^{-8}}{2.25}  = b\\b = 2.77 *10^{-8}

Thus resistance will be R = ρ\frac{L}{A}

where A = π r^2

so,

R = \frac{8.50*10^{-8} * 1.50}{3.14*(1.10*10^{-2})^2} \\R=3.35 * 10 ^{-4}

6 0
1 year ago
A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end. A 2.0-N weight is suspended from
fgiga [73]

Answer:

3.5 N

Explanation:

Let the 0-cm end be the moment point. We know that for the system to be balanced, the total moment about this point must be 0. Let's calculate the moment at each point, in order from 0 to 100cm

- Tension of the string attached at the 0cm end is 0 as moment arm is 0

- 2 N weight suspended from the 10 cm position: 2*10 = 20 Ncm clockwise

- 2 N weight suspended from the 50 cm position: 2*50 = 100 Ncm clockwise

- 1 N stick weight at its center of mass, which is 50 cm position, since the stick is uniform: 1*50 = 50 Ncm clockwise

- 3 N weight suspended from the 60 cm position: 3*60 = 180 Ncm clockwise

- Tension T (N) of the string attached at the 100-cm end: T*100 = 100T Ncm counter-clockwise.

Total Clockwise moment = 20 + 100 + 50 + 180 = 350Ncm

Total counter-clockwise moment = 100T

For this to balance, 100 T = 350

so T = 350 / 100 = 3.5 N

4 0
1 year ago
How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?
podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J

Final velocity = 40 m/s

\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0
2 years ago
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