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
Answer:
The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left.
Explanation:
There are two objects X and Y. Mass of object X is 2M, that is moving with a speed of 
and that of object Y is M that is moving with a speed of 
.
When both of the objects collide, the magnitude of the forces F the objects exert on each other is equal to :
The force exerted by X on Y is F to the right, and the force exerted by Yon X is F to the left using Newton's third law of motion. Hence, the correct option is (c).
A house, address, area.
I hope this helps!
You should just ask the wave
The Virtual Laboratory is an interactive environment for creating and conducting simulated experiments: a playground for experimentation. It consists of domain-dependent simulation programs, experimental units called objects that encompass data files, tools that operate on these objects
