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kipiarov [429]
3 years ago
9

Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force betw

een them?
A): 90 N, repulsive
B): 90 N, attractive
C): 23 N, attractive
D): 45 N, attractive

If one could answer this ASAP, it will be greatly appreciated
Physics
1 answer:
mr Goodwill [35]3 years ago
5 0

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.

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A current of 0. 82 a flows through a light bulb. how much charge passes through the light bulb during 94 s?
gladu [14]

A current of 0. 82A flows through a light bulb. The charge passed through the light bulb during 94 s is 77.08C

The amount of charge flown for a given period of time determines the current passed through a bulb or electrical body.

The relation between the charge, current and time is given as:

Q = I × t

where, Q is the charge flown through bulb

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State coulombs law in word​
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<h2><em>state coulombs law in word</em></h2>

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CISTU U
marusya05 [52]

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

To attain the terminal velocity,

Fiscous force=Weight

or, 23.55× eta = 41.6

or, eta = 1.76

whete eta is the coefficient of viscosity.

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