3 years ago

Prove f=ma i will give

Physics

1 answer:

iragen [17]3 years ago

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Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

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Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

distance between two charges, r = 10 cm = 0.1 m

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

$F = \frac{KQ^2}{r^2}$

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

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In 0.60 seconds, a projectile goes from 0 to 610 m/s. What is the acceleration of the projectile?

IceJOKER [234]

Answer: $a=1016.66 m/s^{2}$

Explanation:

Acceleration $a$ is expressed in the following formula:

$a=\frac{V_{f}-V_{o}}{t}$

Where:

$V_{f}=610 m/s$ is the final velocity of the projectile

$V_{o}=0 m/s$ is the initial velocity of the projectile

$t=0.6 s$ is the time

Solving:

$a=\frac{610 m/s-0 m/s}{0.6 s}$

$a=1016.66 m/s^{2}$ This is the acceleration of the projectile

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Prove f=ma i will give ​

Prove f=ma i will give