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Aleks [24]
3 years ago
9

During a shock which lasts 10ms (0.01s), the voltage difference between the electrodes (so the drop in potential across all thre

e resistors in our model) is 1750V. 200J of energy are dissipated during the shock. What is the average power delivered
Physics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

20000 W

Explanation:

Power: This can be defined as the rate at which energy is dissipated or used. The S.I unit of power is Watt(W).

The expression of power is given as,

P = E/t.............................. Equation 1

Where P = power, E = Energy, t = time.

Given: E = 200 J, t = 0.01 s

Substitute into equation 1

P = 200/0.01

P = 20000 W.

Hence the average power = 20000 W

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Answer:

chloride, ion, -1, this is an anion

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3 years ago
Two Objects of the same size will always have the same mass” Is this statement correct?
attashe74 [19]

No, they won't, mass coincides with density and objects have different densities a one pound lead ball would be smaller than a one pound copper one.

6 0
3 years ago
A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th
Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

C = \epsilon_0 (\frac{A}{d})

Here,

\epsilon_0 =  Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})

C = 1.5901*10^{-10}F

PART A) Energy stored in the capacitor is

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

E = \frac{1}{2} (1.5901*10^{-10})(575)^2

E = 2.628*10^{-5}J

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

Q = CV

Q = (1.59*10^{-10})(575)

Q = 9.1425*10^{-8}C

Now if d = 10.04*10^{-3}m we have that the capacitance is

C = \epsilon_0 (\frac{A}{d})

C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})

C = 3.9754*10^{-11}F

Then the energy stored is

E = \frac{1}{2} \frac{Q^2}{C}

E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})

E = 1.051*10^{-4} J

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

W = 1.051*10^{-4} J -2.628*10^{-5}J

W = 7.882*10^{-5} J

6 0
3 years ago
PLZ HELP MEEEEEEEEE ASAP
mafiozo [28]
The answer is “Impulse acting on it” according to the impulse-momentum theorem.
6 0
3 years ago
The drawing shows Robin Hood (mass = 85.0 kg) about to escape from a dangerous situation. With one hand, he is gripping the rope
mixas84 [53]
The resultant force on the system is equivalent to the difference in the weights of the chandelier and Robin Hood.
F(net) = 240g - 85g
F(net) = 155g

Robin Hood's Acceleration:
F = ma
155g = 85a
a = 17.89 m/s²

Tension = mg + ma
Tension = 85(g + a)
Tension = 2400 N
8 0
3 years ago
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