Ask question

Ask question

All categories

valentina_108 [34]

3 years ago

15

At what distance of separation, r, must two 3.20 x 10-9 Coulomb charges be positioned in order for the repulsive force between t

hem to be 83.4 N?

2 answers:

nataly862011 [7]3 years ago

7 0

F = q₁*q₂* C / r²

F =83.4
q₁ = q₂ = 3.2 * 10⁻⁹
C couloumb constant 8.9 * 10⁹

solve for r

STatiana [176]3 years ago

4 0

I'm not 100% sure but I think the answer is 60.4, because you multiply 3.20 by 10, to get 32 - 9 = 23, then subtract 23 from 83.4. Hope this helps you, and good luck!!!

You might be interested in

In a chemical reaction, one element replaces another element in a compound to form a new substance. Which of the following best

Andreyy89

Answer:

single replacement

Explanation:

In the question it says one element replaces another element which means there is only one replacement.

3 0

3 years ago

2. If you want 0. 250 a (250 milliamps) to flow around a circuit with a resistance of 400 ohms, what voltage do you need?

grigory [225]

Answer:

0.000625 V

Explanation:

The formula linking current , resistance and voltage is :

V = I/R

Voltage = Current / Resistance

Now we substitute values given in question :

Voltage = 0.250 / 400

Voltage (V) = 0.000625

Our final answer is 0.000625 V

Hope this helped and have a good day

5 0

1 year ago

Which statement is true about planets with longer orbits in the solar system?

Yakvenalex [24]

Answer:

D They have longer seasons

Explanation:

I think this because the days will be longer so that means that the seasons will change also.

5 0

3 years ago

Read 2 more answers

A vector points -1.55 units along the x-axis and 3.22 units along the y-axis what is the magnitude of the vector

Rasek [7]

Answer:

3.57 units

Explanation:

$x =\sqrt{ (-1.55)^2+(3.22)^2} = 3.57 units$

5 0

3 years ago

Read 2 more answers

A plane electromagnetic wave, with wavelength 4.1 m, travels in vacuum in the positive direction of an x axis. The electric fiel

marusya05 [52]

(a) $7.32\cdot 10^7 Hz$

The frequency of an electromagnetic waves is given by:

$f=\frac{c}{\lambda}$

where

$c=3.0\cdot 10^8 m/s$ is the speed of light

$\lambda=4.1 m$ is the wavelength of the wave in the problem

Substituting into the equation, we find

$f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz$

(b) $4.60\cdot 10^8 rad/s$

The angular frequency of a wave is given by

$\omega = 2\pi f$

where

f is the frequency

For this wave,

$f=7.32\cdot 10^7 Hz$

So the angular frequency is

$\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s$

(c) $1.53 m^{-1}$

The angular wave number of a wave is given by

$k=\frac{2\pi}{\lambda}$

where

$\lambda$ is the wavelength of the wave

For this wave, we have

$\lambda=4.1 m$

so the angular wave number is

$k=\frac{2\pi}{4.1 m}=1.53 m^{-1}$

(d) $1.03\cdot 10^{-6}T$

For an electromagnetic wave,

$E=cB$

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

$B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T$

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

$I=\frac{E^2}{2\mu_0 c}$

where we have

E = 310 V/m is the amplitude of the electric field

$\mu_0$ is the vacuum permeability

c is the speed of light

Substituting into the formula,

$I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2$

(g) $1.53\cdot 10^{-8} kg m/s$

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

$\frac{dp}{dt}=\frac{A}{c}$

where the <S> is the magnitude of the Poynting vector, given by

$=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2$

and where the surface is

A = 1.8 m^2

Substituting, we find

$\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s$

(h) $8.47\cdot 10^{-7} N/m^2$

For a surface that totally absorbs the wave, the radiation pressure is given by

$p=\frac{}{c}$

where we have

$=254.2 W/m^2$

$c=3\cdot 10^8 m/s$

Substituting, we find

$p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2$

8 0

3 years ago