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1 year ago

One Affective way to develop the feeling function in one’s personality is to

Physics

1 answer:

Mademuasel [1]1 year ago

5 0

Answer:

Explanation:

I think it’s a. Because, I normally think about how someone else feels if I want to say something.

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Describe how the earth is a "giant magnet.

Gala2k [10]

Because the core is a big ball of iron

4 0

3 years ago

Energy slowly leaks outward through the diffusion of photons that repeatedly bounce off ions and electrons

stepan [7]

Energy slowly leaks outward through the radiative diffusion of photons that repeatedly bounce off ions and electrons.

<h3>What is radiative diffusion?</h3>

A radiation zone is a layer of a star's core where energy is mostly carried toward the outside by radiative diffusion and thermal conduction rather than convection.

As photons, energy passes through the radiation zone as electromagnetic radiation.

The radiative diffusion of photons that repeatedly bounce off ions and electrons progressively drains energy outward.

Hence,radiative diffusion is correct answer.

To learn more about radiative diffusion refer:

brainly.com/question/3598352

#SPJ4

7 0

2 years ago

If an electric wire is allowed to produce a magnetic field no larger than that of the Earth (0.503 X 104 T) at a distance of 15

dem82 [27]

Answer:

37.725 A

Explanation:

B = magnitude of the magnetic field produced by the electric wire = 0.503 x 10⁻⁴ T

r = distance from the wire where the magnetic field is noted = 15 cm = 0.15 m

i = magnitude of current flowing through the wire = ?

Magnetic field by a long wire is given as

$B = \frac{\mu _{o}}{4\pi }\frac{2i}{r}$

Inserting the values

$0.503\times 10^{-4} = (10^{-7})\frac{2i}{0.15}$

i = 37.725 A

6 0

2 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

2 years ago

Derive an expression for the gravitational potential energy of a system consisting of Earth and a brick of mass m placed at Eart

Arlecino [84]

Answer:

The gravitational potential energy of a system is -3/2 (GmE)(m)/RE

Explanation:

Given

mE = Mass of Earth

RE = Radius of Earth

G = Gravitational Constant

Let p = The mass density of the earth is

p = M/(4/3πRE³)

p = 3M/4πRE³

Taking for instance,a very thin spherical shell in the earth;

Let r = radius

dr = thickness

Its volume is given by;

dV = 4πr²dr

Since mass = density* volume;

It's mass would be

dm = p * 4πr²dr

The gravitational potential at the center due would equal;

dV = -Gdm/r

Substitute (p * 4πr²dr) for dm

dV = -G(p * 4πr²dr)/r

dV = -G(p * 4πrdr)

The gravitational potential at the center of the earth would equal;

V = ∫dV

V = ∫ -G(p * 4πrdr) {RE,0}

V = -4πGp∫rdr {RE,0}

V = -4πGp (r²/2) {RE,0}

V = -4πGp{RE²/2)

V = -4Gπ * 3M/4πRE³ * RE²/2

V = -3/2 GmE/RE

The gravitational potential energy of the system of the earth and the brick at the center equals

U = Vm

U = -3/2 GmE/RE * m

U = -3/2 (GmE)(m)/RE

5 0

2 years ago