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2 years ago

I'll mark brainliest

Physics

1 answer:

irga5000 [103]2 years ago

7 0

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

= 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

= 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

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If distance between two charges increased by 2 times then force

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Charge Q is distributed uniformly throughout the volume of an insulating sphere of radius R = 4.00 cm. At a distance of r = 8.00

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Answer:

$2.62898\times 10^{-6}\ C/m^3$

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Explanation:

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

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$\sigma=\dfrac{Q}{\dfrac{4}{3}\pi R^3}\\\Rightarrow \sigma=\dfrac{7.04783\times 10^{-10}}{\dfrac{4}{3}\pi (0.04)^3}\\\Rightarrow \sigma=2.62898\times 10^{-6}\ C/m^3$

The volume charge density for the sphere is $2.62898\times 10^{-6}\ C/m^3$

$E=\dfrac{kQr}{R^3}\\\Rightarrow E=\dfrac{8.99\times 10^9\times 7.04783\times 10^{-10}\times 0.02}{0.04^3}\\\Rightarrow E=1979.99974\ N/C$

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2 years ago

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3 years ago

A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro

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Answer:

The correct answer is B

Explanation:

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A = 4π r²

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In this case the entire charge is inside the Gaussian surface, the field is

E = (1 /4πε₀ ) Q / (1.25R)²

E = (1 /4πε₀ ) Q / R2 1 / 1.56²

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C) r = 2R

All charge inside is inside the Gaussian surface

$E_{B}$ =(1 /4π ε₀ ) Q 1/(2R)²

$E_{B}$ = (1 /4π ε₀ ) q/R² 1/4

$E_{B}$ = Eo 1/4

$E_{B}$ = 0.25 Eo

D) False the field changes with distance

The correct answer is B

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