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2 years ago

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A traditional set of cycling rollers has two identical, parallel cylinders in the rear of the device that the rear tire of the b

icycle rests on. Assume that the rear tire is rotating at ω = 32k rad/s. What are the angular velocities of the two cylinders? Consider r1 = 460 mm and r2 = 46 mm.

1 answer:

mars1129 [50]2 years ago

8 0

Answer:

ω2 = 216.47 rad/s

Explanation:

given data

radius r1 = 460 mm

radius r2 = 46 mm

ω = 32k rad/s

solution

we know here that power generated by roller that is

power = T. ω ..............1

power = F × r × ω

and this force of roller on cylinder is equal and opposite force apply by roller

so power transfer equal in every cylinder so

( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2

so

ω2 = $\frac{460\times 32}{34\times 2}$

ω2 = 216.47

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AlekseyPX

Answer:

$P_{out} = 0.100 W = 100 mW$

Explanation:

The attached image shows the system expressed in the question.

We can define an expression for the system.

The equivalent equation for the system would be

$G_{total} = G_{1} + G_{2} + G_{3}\\G_{total} = -16dB+20dB-10 dB = -6 dB$

so, the input signal could be expressed in dB terms

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so the output signal could be expressed as.

$P_{out} = P_{in} + G_{1} + G_{2} + G_{3}\\P_{out} = -3.97 dB - 6dB = -9.97 dB$

The gain should be expressed in dB terms and power in dBm terms so

$P_{out} = -9.97 + 30 = 20.03 dBm$

using the (1) equation to find it in terms of Watts

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2 years ago

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

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Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

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Velocity= 2 * $\frac{65*10^-2}{7.84}$

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So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

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Solving we get

m= 35.58 kg

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2 years ago

The density of nuclear matter is about 1018 kg/m3. Given that 1 mL is equal in volume to 1 cm3, what is the density of nuclear m

Sonbull [250]

Answer:

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Explanation:

given data

density of nuclear = $10^{18}$ kg/m³

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to find out

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solution

we know here

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so

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2 years ago

Read 2 more answers

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Answer:

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$a=5 i+2t j - 6\ t^2k$

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$a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

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b) $a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2$

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2 years ago

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2 years ago