Answer:
+1.46×10¯⁶ C
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = +26.3 μC = +26.3×10¯⁶ C
Force (F) = 0.615 N
Distance apart (r) = 0.750 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Charge 2 (q₂) =?
The value of the second charge can be obtained as follow:
F = Kq₁q₂ / r²
0.615 = 9×10⁹ × 26.3×10¯⁶ × q₂ / 0.750²
0.615 = 236700 × q₂ / 0.5625
Cross multiply
236700 × q₂ = 0.615 × 0.5625
Divide both side by 236700
q₂ = (0.615 × 0.5625) / 236700
q₂ = +1.46×10¯⁶ C
NOTE: The force between them is repulsive as stated from the question. This means that both charge has the same sign. Since the first charge has a positive sign, the second charge also has a positive sign. Thus, the value of the second charge is +1.46×10¯⁶ C
Velocity is speed plus direction, so an example of velocity would be a vehicle traveling at 75mph north. Velocity is a vector quantity because it describes both magnitude and direction.
I would have to see the graph.. but by looking at one one online, they are between points D and E.
Answer:
N = 23.4 N
Explanation:
After reading that long sentence, let's solve the question
The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis
N - w₁ -w₂ =
N = m₁ g + m₂ g
N = g (m₁ + m₂)
let's calculate
N = 9.8 (0.760 + 1.630)
N = 23.4 N
This is the force of the support of the two blocks on the surface.
