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2 years ago

6

Two point charges, initially 3 cm apart, are moved to a distance of 1 cm apart. By what factor does the resulting electric force

between them change?
A. 3
B. 1/9
C. 1/3
D. 9

1 answer:

Solnce55 [7]2 years ago

6 0

The answer is B (1/9).

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What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m

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Answer:

Velocity of a proton, $v=1.7\times 10^6\ m/s$

Explanation:

It is given that,

Potential difference, $V=15\ kV=15\times 10^3\ V$

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

$\dfrac{1}{2}mv^2=qV$

q is the charge of proton

m is the mass of proton

$v=\sqrt{\dfrac{2qV}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}$

$v=1695361.75\ m/s$

$v=1.69\times 10^6\ m/s$

or

$v=1.7\times 10^6\ m/s$

So, the velocity of a proton is $1.7\times 10^6\ m/s$ . Hence, this is the required solution.

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3 years ago

while playing her guitar , karen plucks one string with increasin levels of force. what effect does this have on the sound produ

kirza4 [7]

The sound will get louder.

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3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

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Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

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Assume that the polymer material has a constant refractive index of 1.5. For light of 600nm wavelength at normal incidence, what

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Answer:

Minimum thickness will be 100 nm

Explanation:

We have given refractive index is n = 1.5

Wavelength of the light incidence $\lambda$ = 600 nm

We have to find the smallest thickness of the film so that there will be minimum light reflect

For minimum thickness of non reflecting film

$t=\frac{\lambda }{4n}$ , here t is thickness, $\lambda$ is wavelength and n is refractive index

Putting all values $t=\frac{600}{4\times 1.5}=100nm$

So minimum thickness will be 100 nm

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3 years ago

A hockey puck is sliding at a constant rate of 2m/s on a frictionless surface. How fast will the puck be moving after 10 sec

AnnyKZ [126]

2m/s because the hockey puck is traveling at a constant speed ( acceleration is 0 ). Unless something acts on the hockey puck it will travel 2 m/s forever.

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