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Complete question:
An air-filled parallel-plate capacitor has plates of area 2.90 cm2 separated by 2.50 mm. The capacitor is connected to a(n) 18.0 V battery. Find the value of its capacitance.
Answer:
The value of its capacitance is 1.027 x 10⁻¹² F
Explanation:
Given;
area of the plate, A = 2.9 cm² = 2.9 x 10⁻⁴ m²
separation distance of the plates, d = 2.5 mm = 2.5 x 10⁻³ m
voltage of the battery, V = 18 V
The value of its capacitance is calculated as;

Therefore, the value of its capacitance is 1.027 x 10⁻¹² F
Answer:

Explanation:
The density changes means that the length in the direction of the motion is changed.
Therefore,

Given :
Side, b = h = 0.13 m
Mass, m = 3.3 kg
Density = 8100 
So,


l = 0.024 m
Then for relativistic length contraction,







Therefore, the speed of the observer relative to the cube is 0.9833 c (in the units of c).
You want to use PEMDAS to solve this equation.
4+9*2/3-1
4+18/6-1
4+6-1
10-1
9
Your answer is 9
Thanks -John
If you have anymore question just ask! :)
Answer:
+16 J
Explanation:
We can solve the problem by using the 1st law of thermodynamics:

where
is the change of the internal energy of the system
Q is the heat (positive if supplied to the system, negative if dissipated by the system)
W is the work done (positive if done by the system, negative if done by the surroundings on the system)
In this case we have:
Q = -12 J is the heat dissipated by the system
W = -28 J is the work done ON the system
Substituting into the equation, we find the change in internal energy of the system:
