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
The elastic potential energy of the spring is 0.31 J
Explanation:
The elastic potential energy of a spring is given by
where
k is the spring constant
x is the compression/stretching of the spring
For the spring in this problem, we have:
k = 500 N/m (spring constant)
x = 0.035 m (compression)
Substituting, we find the elastic potential energy:
Learn more about potential energy:
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Chiefly in science fiction) an invisible barrier of exerted strength or impetus.
Answer:
Explanation:
h = Planck's constant = 
c = Speed of light = 
E = Energy = 
Wavelength ejected is given by
The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is
Answer:
32km per hour
Explanation:
Explanation:
In first case v = a t
==> a t = 40 km p h
Now distance covered S1 + S2 + S3
S1 = 1/2 a t^2 and S3 = 1/2 a t^2
But S2 = 3t * 40 = 120 t km
Hence total distance = at^2 + 120 t
Time taken (total) = t + 3t + t = 5 t
Hence average speed = at^2 + 120 t / 5 t
Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour
