E = electric field = 1 x 10³ N/C
q = charge = 4 x 10⁻⁶ C
d = distance = 0.10 m
electric potential energy is given as
U = q E d
inserting the above values
U = (4 x 10⁻⁶ ) (1 x 10³) (0.1)
U = 4 x 10⁻⁴ J
when distance = d = 1.3 m
U = (4 x 10⁻⁶ ) (1 x 10³) (1.3)
U = 5.2 x 10⁻³ J
Answer:
B on Edge 2020
She can change the arrows so they show current traveling in opposite directions on the sides of the loop.
Explanation:
Just took the test haha
B. Resistivity
Resistance offered by a substance of unit area per unit length.
If two particles are involved in an elastic collision, the velocity of the second particle after collision can be expressed as: v2f=2⋅m1(m2+m1)v1i+(m2−m1)(m2+m1)v2i v 2 f = 2 ⋅ m 1 ( m 2 + m 1 ) v 1 i + ( m 2 − m 1 ) ( m 2 + m 1 ) v 2 i .
According to Newton`s law. Force exerted by car,
After adding an additional 400 kg of mass, the force will be same therefore the acceleration
Thus, the acceleration after adding the masses is 1.47 \ m/s^2.