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frosja888 [35]
3 years ago
10

Two particles, one with charge − 7.97 μC −7.97 μC and one with charge 7.75 μC, 7.75 μC, are 8.09 cm 8.09 cm apart. What is the m

agnitude of the force that one particle exerts on the other?
Physics
2 answers:
mr Goodwill [35]3 years ago
4 0

Answer:

Electric force, F = 84.93 N

Explanation:

Given that,

Charge on particle 1, q_1=-7.97\ \mu C=-7.97\times 10^{-6}\ C

Charge on particle 1, q_2=7.75\ \mu C=7.75\times 10^{-6}\ C

The distance between charges, d = 8.09 cm

To find,

The magnitude of the force that one particle exerts on the other.

Solution,

The electric force of attraction or repulsion is given by the formula as:

F=\dfrac{kq_1q_2}{d^2}

k is the electrostatic constant

F=\dfrac{9\times 10^9\times 7.97\times 10^{-6}\times 7.75\times 10^{-6}}{(8.09\times 10^{-2})^2}

F = 84.93 N

So, the magnitude of the force that one particle exerts on the other is 84.93 N.

ZanzabumX [31]3 years ago
3 0

Answer:

84.9 N

Explanation:

q = 7.97 micro coulomb

q' = 7.75 micro coulomb

distance, d = 8.09 cm

The force between the two particles is given by

F=\frac{Kqq'}{r^{2}}

F=\frac{9\times 10^{9}\times 7.97\times 10^{-6}\times 7.75 \times 10^{-6}}{\left (8.09\times 10^{-2}  \right )^{2}}

F = 84.9 N

Thus, the force between the two charges is 84.9 N.

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Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

           -Q is the negative charge

          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

8 0
3 years ago
Water flows through a horizontal pipe. The diameter of the pipe at point b is larger than the diameter of the pipe at point a. W
Natalija [7]

The speed of the water is the greatest at point B

5 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/
Elanso [62]

Given :

Mass of water, m = 2 grams.

The temperature of water drops from 31 °C to 29 °C .

The specific heat of water is 4.184 J/(g • °C).

To Find :

Amount of heat lost in this process.

Solution :

We know, heat lost is given by :

Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J

Therefore, amount of heat lost in this process is 16.736 J.

4 0
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A man standing 1.54 m in front of a shaving mirror produces a real, inverted image 15.2 cm from it. What is the focal length of
zavuch27 [327]

Answer:

The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

Explanation:

Given that,

Object distance u=1.54 m =154 cm

Image distance v = 15.2 cm

Magnification = 2

We need to calculate the focal length

Using formula of mirror

\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{f}=\dfrac{1}{15.2}+\dfrac{1}{-154}

\dfrac{1}{f}=\dfrac{347}{5852}

f=16.86\ cm

We need to calculate the focal length

Using formula of magnification

m= \dfrac{-v}{u}

Put the value into the formula

2=\dfrac{v}{u}

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Using formula of for focal length

\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

\dfrac{1}{16.86}=\dfrac{1}{u}-\dfrac{1}{2u}

\dfrac{1}{16.86}=\dfrac{1}{2u}

2u=16.86

u=\dfrac{16.86}{2}

u=8.43\ cm

Hence, The focal length is 16.86 cm and the distance of the man  if he wants to form an upright image of his chin that is twice the chin's actual size is 8.43 cm.

3 0
3 years ago
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