Answer: q2 = -0.05286
Explanation:
Given that
Charge q1 = - 0.00325C
Electric force F = 48900N
The electric field strength experienced by the charge will be force per unit charge. That is
E = F/q
Substitute F and q into the formula
E = 48900/0.00325
E = 15046153.85 N/C
The value of the repelled second charge will be achieved by using the formula
E = kq/d^2
Where the value of constant
k = 8.99×10^9Nm^2/C^2
d = 5.62m
Substitutes E, d and k into the formula
15046153.85 = 8.99×10^9q/5.62^2
15046153.85 = 284634186.5q
Make q the subject of formula
q2 = 15046153.85/ 28463416.5
q2 = 0.05286
Since they repelled each other, q2 will be negative. Therefore,
q2 = -0.05286
The answer is C. elastic potential energy
ANSWER; KE=5mv^2 so it is proportional to v^2.
Explanation:So if you triple the velocity you are replacing v with 3v. Then you get (3v)^2=9v^2.
Answer:
x = 0.0873 m
Explanation:
given,
length of clothesline = 10 m
the line sags to create an angle that is 1 degree below the horizontal on each end.
As mass is hang at the center the angle made let the deflection be 'x'
as the shirt is hang at the center distance will be equal to 5 m.
now
x = 0.0873 m
Hence, the cloth line will be x = 0.0873 m from the original length of the clothesline.
Answer:
Rf5 is 6 and I have an article on common strategies for coping
Explanation:
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