We need to draw a coordinates. the east and south should be the north vector for horizontal line 50 KM in distance is zero. the south is a negative. the south east and north west we should draw the 45 degrees angle in the approprate quadrant and then use the 1-1-sqrt(2) have a relationship in 45-45-90 triangles to resolve the N/S, E/W components.
hope this help
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
Answer:
a. 0.2 A
Explanation:
Given;
voltage of the battery, V = 12 V
three resistance connected in series, R₁ = 15 Ω, R₂ = 21 Ω, R₃ = 24Ω
The equivalent resistance for series connection is calculated as;
Rt = R₁ + R₂ + R₃
Rt = 15Ω + 21Ω + 24Ω
Rt = 60 Ω
Apply Ohm's law to calculate the current n the circuit;
V = IRt
I = V/Rt
I = 12/60
I = 0.2 A
Therefore, the current in the circuit is 0.2 A
D: An electromagnet with 20 coils.
That is because the more coils of the wire, the stronger the electromagnet is.
True........don't blame me if i'm wrong.<span />
