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lora16 [44]
3 years ago
7

The First Law of Thermodynamics is the same as ______ with heat and work taken into consideration.

Physics
1 answer:
Dafna1 [17]3 years ago
3 0

Answer:

the law of conservation of energy

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An ideal parallel-plate capacitor consists of a set of two parallel plates of area A separated by a very small distance d. When
sleet_krkn [62]

Answer:

2dQ²/2εA = 2U₀

Explanation:

The energy stored in the ideal parallel plate capacitor U = Q²/2C. For a parallel plate capacitor, C = εA/d where A is the area between the plates and D is the distance between them. So

U = Q²/2C = Q²/2εA/d = dQ²/2εA. At distance d₁ = d, U = U₀,

U₀ = dQ²/2εA,

When d₂ = 2d, U₁ = d₂Q²/2εA = 2dQ²/2εA = 2U₀

3 0
3 years ago
Matter is anything that takes up space and has mass. O A. True O B. False​
nadezda [96]

Answer:

A.true

Explanation:

6 0
3 years ago
Read 2 more answers
A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie
-BARSIC- [3]

a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

8 0
3 years ago
The speed of light in a vacuum is approximately 0.3 gm/s. What is the speed of light in meters per second?
klemol [59]

We have to convert Gm/s to m/s.

As  1 \ Gm/s = 10^9 \ m/s

Therefore the speed of light in vacuum,

c = 0.3 \ Gm/s = 0.3 \times 10^9 \ m/s \\\\ c= 3 \times 10^8 \ m/s

Thus, the speed of light in m/s is 3 \times 10^8 \ m/s

7 0
3 years ago
Whta is the mechanical advantage when the effort force is 20 and the resistance force is 5
pogonyaev
The mechanical advantage would be 4. 20/5=4
7 0
3 years ago
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