Multiplying the ideal gas law constant
Answer:
230.4 N
Explanation:
From the question given above, the following data were obtained:
Charge (q) of each protons = 1.6×10¯¹⁹ C
Distance apart (r) = 1×10¯¹⁵ m
Force (F) =?
NOTE: Electric constant (K) = 9×10⁹ Nm²/C²
The force exerted can be obtained as follow:
F = Kq₁q₂ / r²
F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²
F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰
F = 2.304×10¯²⁸ / 1×10¯³⁰
F = 230.4 N
Therefore, the force exerted is 230.4 N
For this problem, we use the conservation of momentum as a solution. Since momentum is mass times velocity, then,
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
where
v₁ and v₂ are initial velocities of cart A and B, respectively
v₁' and v₂' are final velocities of cart A and B, respectively
m₁ and m₂ are masses of cart A and B, respectively
(7 kg)(0 m/s) + (3 kg)(0 m/s) = (7 kg)(v₁') + (3 kg)(6 m/s)
Solving for v₁',
v₁' = -2.57 m/s
<em>Therefore, the speed of cart A is at 2.57 m/s at the direction opposite of cart B.</em>
Answer:
e. The net magnetic flux in this case would be equal to zero.
Explanation:
As per Gauss law of magnetism we need to find the net magnetic flux through a closed loop
here we know that net magnetic flux is the scalar product of magnetic field vector and area vector
so here we have
= net magnetic flux
since we know that magnetic field always forms closed loop so if we find the integral over a closed loop
then in that case the value of the close integral must be zero
so correct answer would be
e. The net magnetic flux in this case would be equal to zero.