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3 years ago

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A student touches sphere x and moves it close to, but not touching sphere y. What are the natures of the charges left on the two

sphere?
Sphere X Sphere Y
A.positive positive
B.no charge no charge
C.positive negative
D.negative positive

1 answer:

e-lub [12.9K]3 years ago

7 0

No charge I know this because

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Any help would be great! Thank you x<br><br><br> Giving brainliest answer xoxo

garri49 [273]

Answer:

A vacuum would have been created. I hope this helps have a great day

3 0

3 years ago

(a) What is the potential between two points situated 10 cm and 20 cm from a 3.0-μC point charge? (b) To what location should th

julia-pushkina [17]

Answer:

(a) 135 kV

(b) The charge chould be moved to infinity

Explanation:

(a)

The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by

$V = -\dfrac{kQ}{r}$

where $k = 9\times 10^9 \text{ F/m}$

Difference in potential between the points is

$kQ\left[-\dfrac{1}{0.2\text{ m}} -\left( -\dfrac{1}{0.1\text{ m}}\right)\right] = \dfrac{kQ}{0.2\text{ m}} = \dfrac{9\times10^9\text{ F/m}\times3\times10^{-6}\text{ C}}{0.2\text{ m}}$

$PD = 135\times 10^3\text{ V} = 135\text{ kV}$

(b)

If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.

$270\times10^3 = kQ\left[-\dfrac{1}{x}-\left(-\dfrac{1}{0.1\text{ m}}\right)\right]$

$10 - \dfrac{1}{x} = \dfrac{270000}{9\times10^9\times3\times10^{-6}} = 10$

$\dfrac{1}{x} = 0$

$x = \infty$

The charge chould be moved to infinity

7 0

3 years ago

PLEASEEE HELPPP!!!!

Zigmanuir [339]

Answer: The work is 1863 N*m

Explanation:

We can define work as:

W = F*d

Where F is the force that the mover needs to apply to the refrigerator, and d is the distance that the refrigerator is moved.

To move the refrigerator, the minimal force that the mover needs to do is exactly the friction force (In this case, the refrigerator will move with constant speed).

Then we will have:

F = 230 N

and the distance is 8.1 meters, then the work will be:

W = 230N*8.1 m = 1863 N*m

3 0

3 years ago

The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr

Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

$I(t)=0.88e^{-t/6\times3600s}$

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

$q=\int\limits {I} \, dt$

Here the limits of integration is from 0 to infinite. So,

$q=\int\limits {0.88e^{-t/6\times3600s}}\, dt$

$q=0.88\times(-6\times3600)(0-1)$

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

$N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}$

N = 1.2 x 10²³

8 0

3 years ago

How to answer this question?

laila [671]

Answer:

measure the vector diagram first

5 0

3 years ago