Answer:
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Explanation:
The formula to be used here is
Q = It
where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C
I is current and it is measured in ampere (amps or A); unknown
t is time and it is measured in seconds (s); 0.05 s
Since, average current is what is unknown
I =Q/t
I = 0.000000028/0.05
I = 5.6 × 10⁻⁷ A
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Answer:
F = -307.4 N
Explanation:
It is given that,
Mass of the baseball, m = 0.145 kg
Initial speed of the baseball, u = 60 m/s
Final speed of the baseball, 
Time of contact, 
(a) It is assumed to find the horizontal component of average force. It is given by :
F = -307.4 N
So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.
Yes,it's true ok? So how have you been doing
Answer:
Explanation:
Due to first charge , electric field at origin will be oriented towards - ve of y axis.
magnitude
Ey = -8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j
= - 31.6 j N/C
Due to second charge electric field at origin
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+ .6²
= 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8
= 18 N/C
It is making angle θ where
Tanθ = .6 / 1.2
= 26.55°
this field in vector form
= - 18 cos 26.55 i - 18 sin26.55 j
= - 16.10 i - 8.04 j
Total field
= - 16.10 i - 8.04 j + ( - 31.6 j )
= -16.1 i - 39.64 j .
Ex = - 16.1 i
Ey = - 39.64 j .