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RUDIKE [14]
2 years ago
9

One way to store energy is in the rotational motion of a flywheel, and some have proposed using such technology to power automob

iles. one unit is based on a 6.5-\(\rm kg\) flywheel in the shape of a hoop of radius 0.11 \(\rm m\) that spins as fast as 70000 \(\rm rpm\).
Physics
1 answer:
miskamm [114]2 years ago
4 0

Answer:

The store energy is 2113109.72 J.

Explanation:

Given that,

Radius = 0.11 m

Mass = 6.5 kg

Spins = 70000 rpm

Suppose we need to find how much kinetic energy is stored by the flywheel when it is rotating at its maximum rate

We need to calculate the moment of inertia

Using formula of moment of inertia

I=mr^2

Put the value into the formula

I=6.5\times(0.11)^2

I=0.07865\ kg-m^2

We need to calculate the rotational kinetic energy

Using formula of rotational kinetic energy

K.E=\dfrac{1}{2}I\omega^2

Put the value into the formula

K.E=\dfrac{1}{2}\times0.07865\times(70000\times\dfrac{2\pi}{60})^2

K.E=2113109.72\ J

Hence, The store energy is 2113109.72 J.

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The components of a 10.8-meters-per-second velocity at an angle of 34.° above the horizontal are
Kisachek [45]

Answer:

6.0 m/s vertical and 9.0 m/s horizontal

Explanation:

For the vertical component, we use the formula:

  • Sin(34°) = <em>y</em> / 10.8

Then we <u>solve for </u><u><em>y</em></u>:

  • 0.559 = <em>y</em> / 10.8
  • <em>y </em>= 6.0

And for the horizontal component, we use the formula:

  • Cos(34°) = <em>x</em> / 10.8

Then we <u>solve for </u><u><em>x</em></u><u>:</u>

  • 0.829 = <em>x</em> / 10.8
  • <em>y </em>= 9.0

So the answer is " 6.0 m/s vertical and 9.0 m/s horizontal".

3 0
2 years ago
Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
schepotkina [342]

Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

3 0
2 years ago
A circuit is made of a battery, a light bulb, and a 2 resistor. The battery has a voltage of 3 volts. When connected, the ammete
Monica [59]

Answer:

3ohms

Explanation:

From Ohm's Law

V = IR

V is that voltage = 3volts

I = current = 1amp

R = resistance in ohms

Putting those values into the above formula.

3volts = 1amp×R

Making R the subject

R = 3/1

R = 3ohms

The resistance of the light bulb is 3ohms.

6 0
3 years ago
Consider this situation: A baseball player dives head-first
siniylev [52]
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.

I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
7 0
2 years ago
An oscilloscope shows a steady sinusoidal signal of 5 Volt peak to peak, which spans 5 cm in vertical direction on the screen. B
Troyanec [42]

Answer:

it will show a continuous rise in value. The rise will be sinusoidal.

Explanation:

3 0
3 years ago
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