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2 years ago

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Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magni

tude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?

1 answer:

IgorLugansk [536]2 years ago

3 0

PDF attached.....................

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You and a partner sit on the floor and stretch out a coiled spring to a length of 7.2 meters. You shake the coil so you

vekshin1

Answer:

Approximately $5.9\; {\rm m\cdot s^{-1}}$ (assuming that the partner is holding the other end of the coil stationary.)

Explanation:

In a standing wave, an antinode is a point that moves with maximal amplitude, while a node is a point that does not move at all. There is an antinode between every two adjacent nodes. Likewise, there is a node between every two adjacent antinodes.

The side of the spring that is being shaken moving with maximal amplitude. Hence, that point on this spring would also be an antinode. In contrast, the side of the spring that is held still (does not move at all) would be a node.

There would be a node between:

the antinode at the end of the spring that is being shaken, and
the antinode between the two ends of this spring.

Overall, the nodes and antinodes on this spring would be:

node at the end that is being held still,
antinode (as mentioned in the question),
node (inferred, not mentioned in the question), and
antinode at the end that is being shaken.

The distance between two adjacent nodes is equal to one-half (that is, $(1/2)$ ) the wavelength of the wave. The distance between a node and an adjacent antinode is one-quarter (that is, $(1/4)$ ) of the wavelength of the wave.

Thus, if the wavelength of the wave in this question is $\lambda$ , the length of this spring would be:

$\displaystyle \frac{1}{2}\, \lambda + \frac{1}{4}\, \lambda = \frac{3}{4}\, \lambda$ .

The question states that the length of this coiled spring is $7.2\; {\rm m}$ . In other words, $(3/4) \, \lambda = 7.2\; {\rm m}$ . The wavelength of this wave would be $(7.2\; {\rm m}) / (3/4) = 9.6\; {\rm m}$ .

The frequency $f$ of this wave is the number of cycles in unit time:

$\begin{aligned} f &= \frac{10}{16.3\; {\rm s}} \approx 0.613\; {\rm s^{-1}}\end{aligned}$ .

Hence, the speed $v$ of this wave would be:

$\begin{aligned} v &= \lambda\, f \\ &=9.6\; {\rm m} \times 0.613\; {\rm s^{-1}} \\ &\approx 5.9\; {\rm m \cdot s^{-1}}\end{aligned}$ .

3 0

1 year ago

The direction of an electric field is always in the direction _______________ would naturally move.

Masja [62]

Answer:

Im not really sure but Id say weather .

Explanation:

6 0

3 years ago

A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .

maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

$F_g = F_c$

$\frac{GmM}{r^2} = \frac{mv^2}{r}$

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

$\frac{GM}{r^2} = \frac{v^2}{r}$

$\frac{GM}{r} = v^2$

$v = \sqrt{\frac{GM}{r}}$

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

$v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}$

$v = 4564.42m/s$

From the speed it is possible to use find the formula, so

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi (6.371*10^6)}{4564.42}$

$T = 8770.05s\approx 146min\approx 2.4hour$

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

$KE = \frac{1}{2} mv^2$

$KE = \frac{1}{2} (100)(4564.42)^2$

$KE = 1.0416*10^9J$

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0

2 years ago

A technician at a semiconductor facility is using an oscilloscope to measure the AC voltage across a resistor in a circuit. The

d1i1m1o1n [39]

Answer:

The value to be reported is 5.48V

Explanation:

The RMS (root mean square) is defined as the value of voltage that will produce the same heating effect, or power dissipation, in circuit, as this AC voltage.

The RMS voltage is also called effective voltage because it is just as effective as DC voltage in providing power to an element.

It is expressed as $V_{rms}$ = $\frac{V_{m} }{\sqrt{2} }$

where Vm is the maximum or peak value of the voltage

In calculating the RMS of the voltage , we simply divide the peak voltage by square root of 2 (√2)

$V_{rms}$ = $\frac{7.75}{\sqrt{2} }$

= $\frac{7.75}{1.414}$

= 5.48 V

6 0

2 years ago

The y component of the electric field of an electromagnetic wave travelling in the +x direction through vacuum obeys the equatio

notsponge [240]

Answer:

λ =8.57 μ m

Explanation:

Given that

Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C

Standard form

Ey=Eo cos[k x-ωt] N/C

By comparing the given equation with the standard wave equation

Eo = 375 N/C

ω = 2.20 × 10¹⁴ rad/s

We know that ω = 2 π f

$f=\dfrac{\omega}{2\pi }$

$f=\dfrac{2.2\times 10^{14}}{2\pi }\ Hz$

f=3.50×10¹³ Hz

We know that the velocity given as

V = f λ

λ =Wavelength

V=Speed = 3 x 10⁸ m/s

$\lambda =\dfrac{V}{f }$

$\lambda =\dfrac{3\times 10^8}{3.5\times 10^{13}}\ m$

λ =0.00000857 m ( 1 μ m = 10⁶ m)

λ =8.57 μ m

8 0

3 years ago