Answer:
1. 8437500 N
2. The force between the two charges is attractive.
Explanation:
1. Determination of the force between the two charges.
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
Distance apart (r) = 80 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 2 × 3 / 80²
F = 5.4×10¹⁰ / 6400
F = 8437500 N
Thus, the force of attraction between the two charges is 8437500 N
2. From the question given, the charges are:
Charge 1 (q₁) = –2.0 C
Charge 2 (q₂) = 3.0 C
We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.
Thus the force between them is attractive.
Answer:
Option D
Explanation:
The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.
In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.
In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.
Answer:
Distance of 400m.
Explanation:
Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).
d = (initial velocity x time) + 1/2 at^2
d = (20 x 10) + 1/2 (4) (10)^2
d = 200 + 200
d = 400 m
Answer:
Explanation:
KE = ½mv² = ½(6.8)8² = 217.6 J
round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.
