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igor_vitrenko [27]
3 years ago
9

Delivery trucks that operate by making use of energy stored in a rotating flywheel have been used in Europe. The trucks are char

ged by using an electric motor to get the flywheel up to its top speed of 940 rad/s. One such flywheel is a solid homogenous cylinder, rotating about its central axis, with a mass of 630 kg and a radius of 1.35 m. What is the kinetic energy of the flywheel after charging?
Physics
1 answer:
Marta_Voda [28]3 years ago
5 0

Answer:

KE = 2.535 x 10⁷ Joules

Explanation:

given,

angular speed of the fly wheel = 940 rad/s

mass of the cylinder = 630 Kg

radius = 1.35 m

KE of flywheel = ?

moment of inertia of the cylinder

I = \dfrac{1}{2}mr^2

 =\dfrac{1}{2}\times 630\times 1.35^2

 = 574 kg m²

kinetic energy of the fly wheel

KE = \dfrac{1}{2}I\omega^2

KE =\dfrac{1}{2}\times 574\times 940^2

KE = 2.535 x 10⁷ Joules

the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules

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<span>The m's cancel μgd = v² / 2 </span>



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<span>d = 8² / 2(0.41)(9.8) </span>

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<span>= 4² / 2(0.46)(9.8) </span>

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Hope i Helped :D
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4 years ago
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A baseball is hit that just goes over a wall that is 45.4m high. If the baseball is traveling at 46.2 m/s at an angle of 32.7° b
mario62 [17]

Answer:

54.9 m/s at 44.9 degrees

Explanation:

If the ball has a total velocity of 46.2 m/s, at an angle of -32.7 degrees, we can decompose its speed into its horizontal and vertical components.

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Vy = V * sin(a) = 46.2 * sin(-32.7) = -25 m/s

SInce there is no force on the horizontal direction (omitting air drag), we can assume constant horizontal speed.

Since a ball thrown is at free fall, only affected by gravity (omitting air drag), we can say it is affected by constant acceleration, therefore we can use

Y(t) = Y0 + Vy0 *t + 1/2 * a * t^2

We consider t=0 as the moment when the ball was hit, so in this case Y0 = 1 m

If we take the first derivative of the equation of position, we get the equation for speed

V(t) = Vy0 + a * t

We know that being t2 the moment the ball goes over the wall

V(t2) = -25 m/s

Y(t2) = 45.4 m

So:

45.4 = 1 + Vy0 * t2 + 1/2 * a * t2^2

-25 = Vy0 + a * t2

Then:

Vy0 = -25 - a * t2

So:

45.4 = 1 + (-25 - a * t2) * t2 + 1/2 * a * t2^2

0 = -44.4 - 25 * t2 - 1/2 * a * t2^2

a = -9.81 m/s^2

0 = -44.4 - 25 * t2 + 4.9 * t2^2

Solving this quadratic equation we get:

t1 = -1.39 s

t2 = 6.5 s

Since we are looking for a positive value we disregard t1.

Now we can obtain Vy0:

Vy0 = -25 + 9.81 * 6.5 = 38.76 m/s

Since horizontal speed is constant Vx0 = 38.9 m/s

By Pythagoras theorem we obtain the value of the initial speed:

V0 = \sqrt{Vx0^2 + Vy0^2} = \sqrt{38.9^2 + 38.76^2} = 54.9 m/s

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a = atan(Vy0/Vx0) = 44.9 degrees

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Does a car have antennae? what senses does it have
Ilia_Sergeevich [38]

Answer:

Explained below.

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Answer:

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E = \frac{\Delta  \phi}{\Delta  t}

\phi  = \texttt {magnetic flux}

E=\frac{\Delta (BA) }{\Delta  t}

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B is the magnetic field strength

substitute

\texttt {substitute}\  \frac{1}{2} \Delta \theta R^2 \ \ for \Delta  A

E=B\frac{(\Delta  \theta R^3/2)}{\Delta  t} \\\\=\frac{1}{2} BR^2\omega

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we plug in the values in the equation above

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E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega

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3 years ago
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