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4 years ago

Mechanical energy is conserved in the presence of which of the following types of forces?magnetic

Physics

1 answer:

Tju [1.3M]4 years ago

7 0

<h2>Answer: electrostatic and gravitational force </h2><h2 />

Mechanical energy remains constant (conserved) if only <u>conservative forces</u> act on the particles.

In this sense, the following forces are conservative:

-Gravitational

-Elastic

-Electrostatics

While the Friction Force and the Magnetic Force are not conservative.

According to this, mechanical energy is conserved in the presence of electrostatic and gravitational forces.

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The Sun delivers an average power of 1150 W/m2 to the top of the Earth’s atmosphere. The permeability of free space is 4π × 10−7

My name is Ann [436]

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed ,E=Electric filed

Power P = I A

A=Area=4πr² ,I=Intensity

$I=\dfrac{CB^2}{2\mu_0}$

$I=\dfrac{CE^2}{2\mu_0 C^2}$

$E=\sqrt{{2I\mu_0 C}}$

$E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}$

E=930.84 N/C

Therefore answer is 930.84 N/C

4 0

4 years ago

A solid sphere of radius 40.0cm has a total positive charge of 26.0μC uniformly distributed throughout its volume. Calculate the

Rudiy27

The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C

R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

$Q\;(\text{total charge of the solid sphere})=(26\;\mathrm{\mu C})\left(\dfrac{1\;\mathrm{C}}{10^6\;\mathrm{\mu C}} \right)={26\times 10^{-6}\;\mathrm{C}}$

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

$E=\dfrac{Q}{4\pi\epsilon_0 r^2}$

Substitute numerical values:

$E&=\dfrac{24\times 10^{-6}}{4\pi (8.8542\times 10^{-12})(0.6)}\\ &={6.49\times 10^5\;\mathrm{N/C}\;\text{directed radially outward}}}$

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).

Learn more about Gaussian sphere here:

brainly.com/question/2004529

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