W = F•dx
F = 65 N,
dx = xf - xi = 0.03 m - (-0.03 m) = 0.06 m
W = 65 N × 0.06 m = 3.9 J
Answer:
1) R1 + ((R2 × R3)/(R2 + R3))
2) 0.5 A
3) 3.6 V
Explanation:
1) We can see that resistors R2 and R3 are in parallel.
Formula for sum of parallel resistors; 1/Rt = 1/R2 + 1/R3
Making Rt the subject gives;
Rt = (R2 × R3)/(R2 + R3)
Now, Resistor R1 is in series with this sum of R2 and R3. Thus;
Total resistance of circuit = R1 + ((R2 × R3)/(R2 + R3))
2) R_total = R1 + ((R2 × R3)/(R2 + R3))
We are given;
R1 = 7.2 Ω
R2 = 8 Ω
R3 = 12 Ω
R_total = 7.2 + ((8 × 12)/(8 + 12))
R_total = 7.2 + 4.8
R_total = 12 Ω
Formula for current is;
I = V/R
I = 6/12
I = 0.5 A
3) since current through the circuit is 0.5 and R1 is 7.2 Ω.
Thus, potential difference through R1 is;
V = IR = 0.5 × 7.2 = 3.6 V
Then every line segment has one and only one mid-point.
（っ＾▿＾）
Answer:
The remaining light bulbs will go out.
Explanation:
The light bulb that was taken out routed power to the other light bulbs, there for, not giving power to the next light bulbs will make them turn off or, "go out". This may be incorrect, as you did not provide a picture of the circuit.
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.
Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg
F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)
F = 83 N
