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3 years ago

12

Which list places the layers of the sun in the correct order from outermost to innermost?

1 answer:

Yuki888 [10]3 years ago

5 0

Is D

is D because the inner layers are the Core, Radiative Zone and Convection Zone.

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Can someone please help me solve these questions

yanalaym [24]

I can't read any of it... Please type a reply comment to this comment, with the questions, and I will help you.

3 0

4 years ago

. How does the intensity of sunlight received affect Earth’s temperatures?

leonid [27]

When the sun's rays strike Earth's surface near the equator, the incoming solar radiation is more direct (nearly perpendicular or closer to a 90˚ angle). Therefore, the solar radiation is concentrated over a smaller surface area, causing warmer temperatures.

7 0

3 years ago

. A 65 kg climber on top of Mt. Everest (8800 m high). How much Pontential Energy does he have?

Kruka [31]

Answer:

He has 5,609,404 J of potential energy

Explanation:

P.E. = M*G*H

65 * 9.8^2 * 8800

7 0

3 years ago

A massless rod is attached to the ceiling by a string. Two weights are hung from the rod: a 0.4-lb weight at its left end and a

user100 [1]

The tension in the string supporting the rod and the attached weights is 7.11 N.

<h3>Tension in the string supporting the rod</h3>

The tension in the string supporting the rod and the attached weights is the sum of the weights supported by the strings.

T = (m1 + m2)g

where;

m1 and m2 are the two masses supported

0.4lb + 1.2lb = 1.6lb = 0.725 kg

T = 0.725 x 9.8

T = 7.11 N

Learn more about tension here: brainly.com/question/918617

#SPJ1

7 0

2 years ago

A solid sphere of radius 30cm is uniformly charged to 100nC. a) What is the volume charge density of the sphere? b) What is the

g100num [7]

Answer

given,

total charge Q = 100 n C

= 100 × 10⁻⁹ C

radius of the solid sphere = 30 cm

= 0.3 m

Volume of sphere = $\dfrac{4}{3}\pi r^3$

= $\dfrac{4}{3}\pi\times 0.3^3$

=0.113 m³

a) volume charge density

$\rho = \dfrac{10^{-7}}{0.133}$

ρ = 8.85 × 10⁻⁷ C/m³

b) at r = 10 cm = 0.1 m

charge in the sphere at radius

$Q = \dfrac{4}{3}\pi\times 0.1^3\time \rho$

= 3.7037 \times 10^{-9}C[/tex]

Field strength

$E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{3.7037 \times 10^{-9}}{4\pi \times 8.85\times 10^{-12}\times 0.1^2}$

= $3.33 \times 10^3 N/C$

at r = 20 cm = 0.2 m

$Q = \dfrac{4}{3}\pi\times r^3\time \rho$

$E_1 = \dfrac{Q}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{ \dfrac{4}{3}\pi\times r^3\time \rho}{4\pi \epsilon_0 r^2}$

$E_1 = \dfrac{\rho}{3 \epsilon_0}$

$E_1 = \dfrac{0.2\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}$

= $6.66 \times 10^3 N/C$

at r = 30 cm

$E_1 = \dfrac{\rho}{3 \epsilon_0}$

$E_1 = \dfrac{0.3\times 8.842 \times 10^{-7}}{3 \times 8.85\times 10^{-12}}$

= 9.99 N/C

6 0

4 years ago