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3 years ago

HELP URGENT!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

nirvana33 [79]3 years ago

5 0

D
Because the rest of the answers are illogical

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How far from a -7.80 μC point charge must a 2.40 μC point charge be placed in order for the electric potential energy of the pai

Naddik [55]

Answer:

Distance between two point charges, r = 0.336 meters

Explanation:

Given that,

Charge 1, $q_1=-7.8\ \mu C=-7.8\times 10^{-6}\ C$

Charge 2, $q_2=2.4\ \mu C=2.4\times 10^{-6}\ C$

Electric potential energy, U = -0.5 J

The electric potential energy at a point r is given by :

$U=k\dfrac{q_1q_2}{r}$

$r=k\dfrac{q_1q_2}{U}$

$r=9\times 10^9\times \dfrac{-7.8\times 10^{-6}\times 2.4\times 10^{-6}}{-0.5}$

r = 0.336 meters

So, the distance between two point charges is 0.336 meters. Hence, this is the required solution.

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3 years ago

A piece of wire length 30cm and cross sectional area of 0.5mm^2 has a resistance of 5ohms at 20°c. It is then heated to a temper

Grace [21]

Answer:

8.333*10^-6 ohms

Explanation:

Resistivity of a material is expressed as;

p = RA/l

R is the resistance of the material

A is the cross sectional area

l is the length of the material

Given

R = 5 ohms

A = 0.5mm^2

A = 5 * 10^-7m^2

l = 30cm = 0.3m

Substitute into the formula;

p = (5 * 5 * 10^-7m^2)/0.3

p = 25 * 10^-7/0.3

p = 0.0000025/0.3

p = 8.333*10^-6

Hence its resistivity at 20 degrees Celsius is 8.333*10^-6 ohms

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3 years ago

A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At

Umnica [9.8K]

Answer:

Explanation:

Given

Charge Q is uniformly spread over large non-conducting Elastic sheet

Electric field due to non-conducting Elastic sheet

$E=\frac{\sigma }{2\epsilon }$

where $\sigma =$ surface charge density $=\frac{q}{d^2}$

$E=\frac{\frac{q}{d^2}}{2\epsilon }$

for side 2d Electric Field is given by

$E'=\frac{\frac{q}{2d^2}}{2\epsilon }$

$E'=\frac{1}{4}\times \frac{\frac{q}{d^2}}{2\epsilon }$

$E'=\frac{E}{4}$

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3 years ago

Nuclear chain reactions within power plants do NOT produce bomb-like explosions primarily because the ________.

Gelneren [198K]

Answer: D. Density of uranium within nuclear fuel rods is insufficient to become explosive

Explanation: Nuclear power plants use the same fuel as nuclear bombs, i.e. radioactive Uranium-235 isotope. However, in a nuclear power plant, the energy is released more slowly unlike in a nuclear bomb. The energy released is through nuclear fission, and radioactive decay occurs at the same rate as in nuclear bombs. therefore, option A, B and C are incorrect.

The primary reason why nuclear chain reactions within power plants do NOT produce bomb-like explosions is because the uranium fuel rods used in electricity generation is not sufficiently enriched in Uranium-235 to produce a nuclear detonation. This is the same idea in option D which is the correct option.

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3 years ago

Formula for percentage error

GarryVolchara [31]

Answer:

PE = (|accepted value – experimental value| \ accepted value) x 100%

Explanation:

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3 years ago