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zimovet [89]
3 years ago
13

Formula for calculating work done in machines ​

Physics
1 answer:
Archy [21]3 years ago
7 0

Answer:

force×distance

Explanation:

i think that's the answer and it could also be expressed as m×g×d

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Negatively charged particle that orbits the nucleus
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Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

If this helped please mark me brainliest.

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Compare and contrast camera obscura with what you know about modern digital photography, including cell phones.
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Answer:

It captures images but does not preserve them.

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An experiment is performed on an unknown material and produces the given heat curve. The temperature of the material is shown as
BlackZzzverrR [31]

Answer:

In the table, 1=46.7 °C, 1=165 J, 2=819 J, 3=1510 J, and 4=2830 J.

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If the sample of material has a mass of =8.60 g, calculate the specific heat when this material is a solid, and when it is liquid, l

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2 years ago
A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
Rama09 [41]

1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

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

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

While the separation between the plates is

d=0.50 mm=5\cdot 10^{-4} m

So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

And now we can find the energy stored,which is given by:

U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

The magnitude of the electric field is given by

E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

E=4\cdot 10^5 V/m, we find

u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3

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