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3 years ago

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Trucks can be run on energy stored in a rotating flywheel, with an electric motor getting the flywheel up to its top speed of 20

0π rad/s. Suppose that one such flywheel is a solid, uniform cylinder with a mass of 500 kg and a radius of 1.0 m. (a) What is the kinetic energy of the flywheel after charging? (b) If the truck uses an average power of 8.0 kW, for how many minutes can it operate between chargings?

1 answer:

Genrish500 [490]3 years ago

3 0

Answer:

The time it can operate between chargins in minutes is

$t=102.8 minutes$

Explanation:

Given: $m=500kg$ , $r=1.0m$ , $w=200\pi rad/s$

a). The rotational kinetic energy

$K_R=\frac{1}{2}*I*w^2$

$I=\frac{1}{2}*m*r^2$

$I=\frac{1}{2}*500kg*(1.0m)^2$

$I=250 kg*m^2$

$K_R=\frac{1}{2}*250kg*m^2*(200\pi rad/s)^2$

$K_R=49.348x10^6J$

b). The power average 0.8kW un range time can be find

$P=\frac{K_R}{t'}$

Solve to t'

$t=\frac{K_R}{P}$

$t=\frac{49.348x10^6}{0.8x10^3w}=6168.5s$

$t=6168.5s\frac{1minute}{60s}=102.8 minutes$

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B. 439nm

C. 59.5 angstroms

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$r_5=5^2(2.380)=59.5$

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9. Consider the elbow to be flexed at 90 degrees with the forearm parallel to the ground and the upper arm perpendicular to the

mojhsa [17]

Answer:

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Explanation:

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∑ τ = I α

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They indicate the angular acceleration, for which we must look for the moments of inertia of the elements involved

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∑ τ = I α

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the moment of inertia of the mass in the hand, let's approach as punctual

I_mass = m L²

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∑ τ = (⅓ 2.3 0.5² + 0.5 0.5²) 10

∑ τ = 3.17 N / m

Moment with respect to ELBOW

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Στ = I α

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