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Angelina_Jolie [31]

3 years ago

15

1. li took 2 } seconds for a car's vclocity to change from 20 m/s to 15 m/s. The mass of the car was 1370 kg. What force was req

uired to cause the acceleration? (Hint: First calculate the acceleration.

1 answer:

Pani-rosa [81]3 years ago

3 0

Answer:

Do you still need help??

Explanation:

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Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons

yarga [219]

Answer:

$q\approx 6.6\cdot 10^{13}~electrons$

Explanation:

<u>Coulomb's Law</u>

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

$\displaystyle F=k\frac{q^2}{d^2}$

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

$\displaystyle q=\sqrt{\frac{F}{k}}\cdot d$

Substituting values:

$\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1$

$q = 1.05\cdot 10^{-5}~c$

This charge corresponds to a number of electrons given by the elementary charge of the electron:

$q_e=1.6 \cdot 10^{-19}~c$

Thus, the charge of any of the spheres is:

$\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}$

$\mathbf{q\approx 6.6\cdot 10^{13}~electrons}$

5 0

2 years ago