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Liula [17]
3 years ago
11

Problem (2) A 16 kg cylinder, initially at rest, is held by a cord connected to a grooved drum whose mass is 20 kg. The drum has

an outer radius �% = 250 mm and an inner radius of �& = 160 mm. If the drum experiences a constant frictional moment of 3 N∙m at O, how far has the cylinder dropped when it has a downward velocity of 2 m/s? Neglect the mass of the cord and treat the drum as a thin disk. Use the Principle of Work and Energy.

Physics
1 answer:
Mice21 [21]3 years ago
4 0

Answer:

Explanation:

The solution to the problem is given in the pictures attached below; the three pictures explains the problem fully and I hope it helps you. Thank you

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State the conservation of momentum theorem
Luda [366]

Answer:

The total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

7 0
3 years ago
How much work, in Joules. is done when a pressure of 95.000 Pa changes a volume from 15 ma to 43 m2?
gayaneshka [121]

Answer:

-2.66 kJ

Explanation:

At constant pressure, work done is:

W = -P ΔV

W = -(95.000 Pa) (43 m² − 15 m²)

W = -2660 J

W = -2.66 kJ

5 0
4 years ago
When one form of energy is converted into another what is the name of that
Svetradugi [14.3K]
Transformation of energy, because energy cannot be created or destroyed
8 0
3 years ago
The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate
Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

L = 10log_{10}\frac{I}{I_{o}}          

I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}      

L' = 10log_{10}\frac{I'}{I_{o}}

I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}        

where

I_{o} = 10^{- 12} W/m^{2}

I = Intensity of sound

Now,

I = \frac{P}{4\pi R^{2}}

Similarly,

I' = \frac{P}{4\pi (R + 11.0)^{2}}

Now,

\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}

\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}

R ^{2} + 22R + 121 = 11.64R^{2}}

10.64R ^{2} - 22R - 121 = 0

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0
3 years ago
Voltage needed to raise current to 3.75a using 20,20,200 resistor set
Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

We are given:

R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega

Using above equation, we get:

\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega

Calculating the voltage by using Ohm's law:

V=IR         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = 9.52\Omega

Putting values in equation 1, we get:

V=3.75\times 9.52\\\\V=35.7V

Hence, the voltage needed is 35.7 V

7 0
3 years ago
Read 2 more answers
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