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3 years ago

Hurry help please, number 3

Physics

2 answers:

STatiana [176]3 years ago

8 0

C) Most of earths weather occurs in this layer.

Dovator [93]3 years ago

6 0

C most of the Earths weather occurs in this layer.

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Rewrite the following sentence in the negative form >Each library contains 3000 brand new books

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2 years ago

In nuclear fission and fusion the mass defect is

riadik2000 [5.3K]

Option C

In nuclear fission and fusion the mass defect is the mass lost during the reaction that is converted into energy

<u>Explanation:</u>

Mass defect is the contrast within the estimated mass of the released system and the empirically estimated mass of the nucleus. The nuclear binding energy is acknowledged as mass, and that mass enhances "missing".

This missing mass is described as a mass defect, which is nuclear energy, also acknowledged as the mass discharged from the reaction as any trajectories. The mass defect of a nucleus depicts the mass of the energy adhesive of the nucleus and is the variation amidst the mass of a nucleus and the entirety of the masses of the nucleons of which it is comprised.

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4 years ago

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What are scientific theories?

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A scientific theory is an explanation of an aspect of the natural world that can be repeatedly tested and verified in accordance with the scientific method, using accepted protocols of observation, measurement, and evaluation of results. Where possible, theories are tested under controlled conditions in an experiment.

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3 years ago

Sail's size and material can make a difference in the effectiveness of how it works because?

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3 years ago

A charge of −20 µC is distributed uniformly over the surface of a spherical conductor of radius 11.0 cm. Determine the electric

Alex73 [517]

Answer:

(a) $-6.76\times 10^{12}\ N/C$

(b) $-1.352\times 10^{13}\ N/C$

(c) $-7.2\times 10^{11}\ N/C$

Explanation:

(a)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 5 cm = 0.05 m

Coulomb's constant (k) = $9\times 10^{9}\ Nm^2/C^2$

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.05\\\\E_{in}=-1.352\times 10^{14}\times 0.05\\\\E_{in}=-6.76\times 10^{12}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(b)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 10 cm = 0.10 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r ≤ R' from the center of sphere is given as:

$E=(\frac{kQ}{R^3})r$

Plug in the given values and solve for 'E'. This gives,

$E_{in}=(\frac{9\times 10^{9}\times -20}{(0.110)^3})\times 0.10\\\\E_{in}=-1.352\times 10^{14}\times 0.10\\\\E_{in}=-1.352\times 10^{13}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

(c)

Given:

Charge on sphere (Q) = $-20\ \mu C=-20\times 10^{-6}\ C$

Radius of sphere (R) = 11.0 cm = 0.110 m

Distance from the center (r) = 50 cm = 0.50 m

Now, we know from Gaussian law for uniform charged spheres, the electric field at a distance 'r > R' from the center of sphere is given as:

$E=\dfrac{kQ}{r^2}$

Plug in the given values and solve for 'E'. This gives,

$E_{out}=(\frac{9\times 10^{9}\times -20}{(0.50)^2})\\\\E_{out}=-7.2\times 10^{11}\ N/C(Negative\ sign\ implies\ radially\ inward\ direction)$

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4 years ago