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3 years ago

What will happen to the force felt between two charged objects if the distance between them is 1/3rd of the original distance

Physics

1 answer:

kenny6666 [7]3 years ago

4 0

Answer:

New force = 9(initial force)

Explanation:

The force between two charges is given by :

$F=\dfrac{kq_1q_2}{r^2}$

Where

d is the original distance

Let d' is the new distance such that, r' = r/3

New force,

$F'=\dfrac{kq_1q_2}{r'^2}\\\\F'=\dfrac{kq_1q_2}{(\dfrac{r}{3})^2}\\\\F'=9\times \dfrac{kq_1q_2}{r^2}\\\\F'=9F$

So, the new force becomes 9 times the initial force.

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How long is the racetrack if it takes a racecar 3.4s traveling at 75 m/s to finish?

ollegr [7]

Answer:127.5m

Explanation:

S=? T=3.4s v= 75 m/s. U=0

S=( u + v)/2 x t

S = 0 +75/2

= 37.5 x 3.4

S= 127.5 m

7 0

3 years ago

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

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3 years ago

Where is the distribution of positive charges found in the atom?

Agata [3.3K]

I think the answer is proton

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4 years ago

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