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Ostrovityanka [42]
2 years ago
11

Consider a uniformly charged sphere of total charge Q and radius R centered at the origin. We want to find the electric field in

side the sphere (r
Physics
1 answer:
laiz [17]2 years ago
8 0

Answer:

Hello your question is incomplete attached below is the complete question

answer :

Total charge enclosed within the sphere : \frac{q_{r1} }{4\pi e_{0}R^3 } . r

Total charge enclosed outside the sphere : \frac{q}{4\pi e_{0}r^2 } .r

Explanation:

Given data:

Total charge of a uniformly charged sphere = Q

radius = R

first step : find the electric field inside and outside the uniformly charged sphere

2nd step : determine the total charge enclosed within and outside the sphere

make a sketch of the uniformly charged sphere

Attached below is a detailed solution

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A speedboat initially at rest accelerates at 4.0 m/s (squared) for 7.0 s. How far does the speedboat move in 7.0 s?
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I think is 3 becuase when he star at 4
4 0
2 years ago
According to the graph, how many atoms would remain after two half-lives?
Fudgin [204]

Answer:

Let No be initial no of atoms

N = N0 / 2      after 1 half-life

N = N0 / 4     after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

4 0
2 years ago
An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
just olya [345]

Answer:

2.78\times 10^{-35}\ \text{kg m/s}

6.178\times 10^{-34}\ \text{m/s}

0.31\times 10^{-4}\ \text{m/s}

Explanation:

\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

The minimum uncertainty in the object's velocity is 0.31\times 10^{-4}\ \text{m/s}.

6 0
2 years ago
If a 25 kg object is moving at a velocity of 10 m/s, the object has<br> energy. Calculate it.
Paul [167]

Answer: Your answer is 1250J

Explanation:

K E = 1/2 m v 2

The mass is  

m = 25 k g

The velocity is  v = 10 m s − 1

So,

K E = 1 /2 x25 x 10 2^2= 1250 J

pls mark brainiest answer  

4 0
2 years ago
Give reasons.<br>a) Sunlight does not fall on the ground where cloud forms shadow.​
Viefleur [7K]

Answer:

Sunlight passing through a cloud is scattered in various directions. Some photons eventually find their way out the bottom of the cloud, some are reflected upwards, and some are absorbed, serving to warm the cloud. Since only a fraction of the light makes it to ground level, you have a shadow.

Explanation:

Hope it helps you

4 0
3 years ago
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