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levacccp [35]
2 years ago
9

Jaiden is writing a report about the structure of the atom. In her report, she says that the atom has three main parts and two s

ubatomic particles. Do you agree with her? Why or why not?
Physics
2 answers:
USPshnik [31]2 years ago
7 0
No because an atom consists of <u>two</u> main parts <em>and</em> <u>three</u> subatomic particles - protons, neutrons, electrons. Each one is smaller than an atom, therefore they are subatomic particles. An atom only requires protons and electrons to be an atom - e.g. Hydrogen has 1 proton and 1 electron. Neutrons do not affect the overall charge of the atom, and only increase the atomic mass.
Anna71 [15]2 years ago
3 0

I disagree with her because an atom has two main parts: the nucleus and the electron cloud. Atoms have three subatomic particles: protons, neutrons, and electrons.



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Determine the centripetal force upon a 40-kg child who makes 10 revolution around the cliffhanger in 29.3 seconds.the radius of
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Answer:

The centripetal force acting on the child is 39400.56 N.

Explanation:

Given:

Mass of the child is, m=40\ kg

Radius of the barrel is, R=2.90\ m

Number of revolutions are, n =10

Time taken for 10 revolutions is, t=29.3\ s

Therefore, the time period of the child is given as:

T=\frac{n}{t}=\frac{10}{29.3}=0.341\ s

Now, angular velocity is related to time period as:

\omega=\frac{2\pi}{T}=\frac{2\pi}{0.341}=18.43\ rad/s

Now, centripetal force acting on the child is given as:

F_{c}=m\omega^2 R\\F_{c}=40\times (18.43)^2\times 2.90\\F_{c}=40\times 339.66\times 2.90\\F_{c}=39400.56\ N

Therefore, the centripetal force acting on the child is 39400.56 N.

8 0
2 years ago
The inductor in a radio receiver carries a current of amplitude 200 mA when a voltage of amplitude 2.40 V is across it at a freq
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Answer:92

Explanation:

3 0
2 years ago
A metallic sphere of radius 2.0 cm is charged with +5.0-μC+5.0-μC charge, which spreads on the surface of the sphere uniformly.
sladkih [1.3K]

Answer:

Explanation:

Potential due to a charged metallic sphere having charge Q and radius r on its surface will be

v = k Q / r . On the surface and inside the metallic sphere , potential is the same . Outside the sphere , at a distance R from the centre  potential is

v = k Q / R

a ) On the surface of the shell , potential due to positive charge is

V₁ = \frac{9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

On the surface of the shell , potential due to negative  charge is

V₁ = \frac{- 9\times10^9\times5\times10^{-6}}{6\times10^{-2}}

Total potential will be zero . they will cancel each other.

b ) On the surface of the sphere potential

= \frac{9\times10^9\times5\times10^{-6}}{2\times10^{-2}}

= 22.5 x 10⁵ V

On the surface of the sphere potential due to outer shell

= \frac{9\times10^9\times5\times10^{-6}}{5\times10^{-2}}

= -9 x 10⁵

Total potential

=( 22.5 - 9 ) x 10⁵

= 13.5 x 10⁵ V

c ) In the space between the two , potential will depend upon the distance of the point from the common centre .

d ) Inside the sphere , potential will be same as that on the surface that is

13.5 x 10⁵ V.

e ) Outside the shell , potential due to both positive and negative charge will cancel each other so it will be zero.

5 0
2 years ago
a film of transparent material 120 nm thick and having refractive index 1.25 is placed on a glass sheet having refractive index
Eva8 [605]

Answer:

a) 600nm

b) 300nm

Explanation:

the path difference = 2t  

t = thickness of the film

L' = wavelength of light in film = L/n

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n = refractive index of glass

(a)

for destructive interference 2t = L'/2 = L/2n

L = 4*t*n

= 4*120*10^-9*1.25  

L = 600 nm

(b)

for constructive interference 2t = L' = L/1.25

L = 2tn

= 2 × 1.25 ×  120nm

= 300 nm

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KiRa [710]
C I believe is the correct answer. Developing possible solutions would be easier than spending hours researching or identifying the need.
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