"Put the moon phases in order from first to last phase.

A particle with charge +4.20 nC is in a uniform electric field E directed to the left. The charge is released from rest and move

Sonja [21]

Answer:

The work done by the electric force is = $+2.20*10^{-6} J$

The potential of the starting point with respect to the end point is $= 523.8 V$

The magnitude of E $= 8730 \ N/C$

Explanation:

Looking at this we can see that the work done by the electric force is equal to the kinetic energy this is because on the electric field is uniform and that the work done resulted in motion.

Mathematically

$\Delta V q = W$

Where $\Delta V$ is the potential difference

q is the charge whose value i given as = $4.20 *10 ^{-9} C$

W is the workdone whose value is obtained as = $+2.20*10^{-6} J$

Substituting

$\Delta V = \frac{W}{q}$

$=\frac{2.2*10^{-6}}{4,20*10^{-9}}$

$= 523.8 V$

Mathematically

$\Delta V = Ed$

Where E is the electric field strength

d is the distanced moved and it is given as 6.00 cm = 0.06 m

Making E the subject of the formula

$E = \frac{\Delta V}{d}$

$E = \frac{523.8}{0.06} = 8730 \ N/C$

3 0

4 years ago

Explain energy transformation using basketball

crimeas [40]

Multiple choice? or no

3 0

3 years ago

A violin string has a length of 327mm and produces a note of frequency 440Hz.

Scorpion4ik [409]

The characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is:

The new frequency is f = 657 Hz

<h3>How is a standing wave produced?</h3>

A standing wave is produced when a traveling wave meets an obstacle and bounces, the sum of the two waves results in a wave that does not propagate in space.

In the event that the obstacle is a fixed point, there is a node at this point. The expression for the length of the standing wave.

L = $\frac{\lambda }{2}$ fundamental frequency

L = $2 \frac{\lambda}{2}$ second harmonic

L = $3 \frac{\lambda}{2}$ third harmonic

L = $n \frac{\lambda}{2}$ general term.

Where L is the length of the chord, lan the wavelength and n an integer.

Wave speed is related to wavelength and frequency.

v = λ f.

Let's substitute.

v = $\frac{2L}{n}$

They indicate that initially the string has a length of L₀ = 327 mm= 0.327m and the frequency is f₀ = 440 Hz.

v n = 2L₀ f₀

v n = 2 0.327 440

v n = 287.76

They indicate that the tension on the string do not changes and the speed of the wave depends only on the tension and the density of the string, therefore it is constant, we assume that the harmonic does not change either, therefore the new length.

v n = 2 L f

Let's substitute.

287.76 = 2 L f

f = $\frac{287.76x}{2L}$

Let's calculate.

f = $\frac{287.76}{2 \ 0.219}$

f = 656.99 Hz

In conclusion with the characteristics of the standing wave we can find the backlash for the frequency of the wave when the string is shortened is: