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3 years ago

Energy that vibrates electrons causing electricity

1 answer:

4 0

Answer: Electromagnetic waves are generated by moving electrons. An electron generates an electric field which we can visualize as lines radiating from the electron Figure 10a. If the electron moves, say it vibrates back and forth, then this motion will be transferred to the field lines and they will become wavy Figure 10b.

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A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3.0 m is initially at rest. A 20 kg boy approaches the m

nekit [7.7K]

Answer:

The velocity of the merry-go-round after the boy hops on the merry-go-round is 1.5 m/s

Explanation:

The rotational inertia of the merry-go-round = 600 kg·m²

The radius of the merry-go-round = 3.0 m

The mass of the boy = 20 kg

The speed with which the boy approaches the merry-go-round = 5.0 m/s

$F_T \cdot r = I \cdot \alpha = m \cdot r^2 \cdot \alpha$

Where;

$F_T$ = The tangential force

I = The rotational inertia

m = The mass

α = The angular acceleration

r = The radius of the merry-go-round

For the merry go round, we have;

$I_m \cdot \alpha_m = I_m \cdot \dfrac{v_m}{r \cdot t}$

$I_m$ = The rotational inertia of the merry-go-round

$\alpha _m$ = The angular acceleration of the merry-go-round

$v _m$ = The linear velocity of the merry-go-round

t = The time of motion

For the boy, we have;

$I_b \cdot \alpha_b = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Where;

$I_b$ = The rotational inertia of the boy

$\alpha _b$ = The angular acceleration of the boy

$v _b$ = The linear velocity of the boy

t = The time of motion

When the boy jumps on the merry-go-round, we have;

$I_m \cdot \dfrac{v_m}{r \cdot t} = m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t}$

Which gives;

$v_m = \dfrac{m_b \cdot r^2 \cdot \dfrac{v_b}{r \cdot t} \cdot r \cdot t}{I_m} = \dfrac{m_b \cdot r^2 \cdot v_b}{I_m}$

From which we have;

$v_m = \dfrac{20 \times 3^2 \times 5}{600} = 1.5$

The velocity of the merry-go-round, $v_m$ , after the boy hops on the merry-go-round = 1.5 m/s.

5 0

3 years ago

Calculate the terminal velocity of

adell [148]

I’m pretty sure it’s a

6 0

3 years ago

Fg =G m1m2/r2 solver for G

Blababa [14]

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Explanation:

Solving for G simply implies that we make G the subject of the formula:

Given equation:

F $_{g}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$

To make G the subject of this expression follow these steps:

Multiply both sides of the equation by $r^{2}$

F $_{g}$ x $r^{2}$ = G $\frac{m_{1} m_{2} }{ r^{2} }$ x $r^{2}$

This gives:

F $_{g}$ $r^{2}$ = G $m_{1} m_{2}$

Multiply both sides by $\frac{1}{m_{1} m_{2} }$

F $_{g}$ $r^{2}$ x $\frac{1}{m_{1} m_{2} }$ = $\frac{1}{m_{1} m_{2} }$ x G $m_{1} m_{2}$

Therefore:

G = $\frac{F_{g} r^{2} }{m_{1} m_{2} }$

Learn more:

Solving formula brainly.com/question/2998489

#learnwithBrainly

6 0

3 years ago

~Correctly~ PLEASE HURRY! :)

suter [353]

A) focal point is the correct awnser

7 0

3 years ago

Read 2 more answers

After your school's team wins the regional championship, students go to the dorm roof and start setting off fireworks rockets. T

Nataliya [291]

Answer:

174m

Explanation:

We can calculate the power of the sound:

$P = IA = I \pi R^2 = 1.52\times10^{-6}*\pi*123^2 = 0.0722 W$

For the intensity to be half of I, then the cover area must be

$a = P/(I/2) = 2P/I = \frac{2*0.722}{1.52\times10^{-6}} = 95058 m^2$

Then the distance from the source center to this point is

$a = \pi r^2$

$r^2 = a / \pi = 95058 / \pi = 30258$

$r = \sqrt{30258} = 174 m$

4 0

3 years ago

Read 2 more answers