Answer: Newton's third law
Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction. The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object.
Explanation:
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
Answer:
In the case of an electric bulb, the electrical energy is converted to light and heat. The amount of electrical energy put into a bulb = the amount of light energy (desirable form) plus the heat energy that comes out of the bulb (undesirable form).
Explanation:
sana nakatulong)):
Answer:
2.5 mi/s^2
Explanation:
please see paper for work!
Answer:
Period of oscillation, T = 2 sec
Explanation:
It is given that,
Mass of the object, m = 5 kg
Spring constant of the spring, k = 50 N/m
This object is hanging from a coiled spring. We need to find the period of oscillation of the spring. The time period of oscillation of the spring is given by :
T = 1.98 sec
or
T = 2 sec
So, the period of oscillation is about 2 seconds. Hence, this is the required solution.
