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damaskus [11]
3 years ago
5

Two spheres carry a charge of 1x10^-7 C. Their centers are separated by 0.7m. Calculate the magnitude of the force between the t

wo charged objects.
Write your answer in scientific notation using one significant figure

PLEASE HELP ME
Physics
1 answer:
Drupady [299]3 years ago
8 0

We have that the  force between  the two charged objects is mathematically given as

F=2.349N

<h3>Magnitude of the force between  two charged objects.</h3>

Question Parameters:

Two spheres carry a charge of 1x10^-7 C.

Their centers are <u>separated</u> by 0.7m.

Generally the equation for the Force  is mathematically given as

F=\frac{q1q2}{4*\pi e*r}

Therefore

F=\frac{(1e-7)^2}{9e9*0.7}

F=2.349N

For more information on Force visit

brainly.com/question/26115859

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A 1,70 atm, una muestra de gas ocupa 4,25 litros. Si la presión en el gas aumenta a 2.40 atm, ¿cuál será el nuevo volumen?
statuscvo [17]

At 1.70 atm, a gas sample occupies 4.25 liters. If the pressure in the gas increases to 2.40 atm, what will the new volume be?

Answer:

3.01L

Explanation:

Given parameters:

Initial pressure, P1  = 1.7atm

Initial volume, V1  = 4.25L

Final pressure, P2  = 2.4atm

Unknown:

Final or new volume, V2  = ?

Solution:

To solve this problem, we use Boyle's law which states that "the volume of a fixed mass of a gas varies inversely as the pressure changes, if the temperature is constant".

            P1 V1  = P2 V2

P1 is the initial pressure

V1 is the initial volume

P2 final pressure

V2 final volume

        1.7 x 4.25  = 2.4 x V2

             V2  = 3.01L

8 0
3 years ago
The speed of a wave in a medium is effected by____,____, and____
lisov135 [29]

\huge\mathfrak\green{answer}

As per as my knowledge

The speed of a wave in a medium is affected by <u>d</u><u>e</u><u>n</u><u>s</u><u>i</u><u>t</u><u>y</u>,<u> </u><u>w</u><u>a</u><u>v</u><u>e</u><u>l</u><u>e</u><u>n</u><u>g</u><u>t</u><u>h</u> and <u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u>:)

(Good luck on your test and mark me brainliest if this helps)

7 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
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