Answer:
f = 485.62 N
Explanation:
Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:
Unbalanced Force = Horizontal Component Applied Force - Frictional Force
Unbalanced Force = Fx - f
But, from Newtons Second Law of Motion:
Unbalanced Force = ma
comparing the equations:
ma = Fx - f
f = F Cos θ - ma
where,
f = frictional force = ?
F = Applied force = 593 N
m = mass of person = 49 kg
a = acceleration = 0.57 m/s²
θ = Angle with horizontal = 30°
Therefore,
f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)
f = 513.55 N - 27.93 N
<u>f = 485.62 N</u>
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Answer:
1.) 274.5v
2.) 206.8v
Explanation:
1.) Given that In one part of the lab activities, students connected a 2.50 µF capacitor to a 746 V power source, whilst connected a second 6.80 µF capacitor to a 562 V source.
The potential difference and charge across EACH capacitor will be
V = Voe
Where Vo = initial voltage
e = natural logarithm = 2.718
For the first capacitor 2.50 µF,
V = Vo × 2.718
746 = Vo × 2.718
Vo = 746/2.718
Vo = 274.5v
To calculate the charge, use the below formula.
Q = CV
Q = 2.5 × 10^-6 × 274.5
Q = 6.86 × 10^-4 C
For the second capacitor 6.80 µF
V = Voe
562 = Vo × 2.718
Vo = 562/2.718
Vo = 206.77v
The charge on it will be
Q = CV
Q = 6.8 × 10^-6 × 206.77
Q = 1.41 × 10^-3 C
B.) Using the formula V = Voe again
165 = Vo × 2.718
Vo = 165 /2.718
Vo = 60.71v
Q = C × 60.71
Q = C
