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Yanka [14]
3 years ago
13

Which layer of the sun is seen during a total solar eclipse?

Physics
1 answer:
maks197457 [2]3 years ago
4 0

Answer:

The Corona.

Explanation:

The Corona is the outermost part of the Sun. Usually, it is nto visible due to the light emitted by the Sun. During a total solar eclipse, that is the moon comes between the Earth and the Sun, the glowing white edge around the eclipse is the Corona.

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According to the law of conservation of energy,
vodka [1.7K]

According to the law of conservation of energy,  

A. an object loses most of its energy as friction

<u>B. the total amount of energy for a system stays the same</u>

Energy is never lost due to the law of conservation

C. the potential energy of an object is always greater than its kinetic energy

D. the kinetic energy of an object is always greater than its potential energy​

8 0
3 years ago
Read 2 more answers
Block 1 (mass 2.00 kg) is moving rightward at 10.0 m/s and block 2 (mass 5.00 kg) is moving rightward at 3.00 m/s. The surface i
DaniilM [7]

Answer:

a) 0.25m

b) 5 m/s

Explanation:

When the spring is compressed both boxes are moving with the same velocity, so applying the principle of linear momentum conservation:

m1*v_{o1}+m2*v_{o2}=(m1+m2)*v\\v=5m/s

Now applying the principle of energy conservation:

K1+K2+U_{g1}-U_e=Kf+U_{g2}\\K1+0-U_e=K2+0\\U_e=K1+K2-kf\\\frac{1}{2}*k*x^2+=\frac{1}{2}*m1*v1^2+\frac{1}{2}*m1*v1^2-\frac{1}{2}*(m1+m2)*v^2\\\\x=\sqrt{\frac{2.00kg*(10m/s)^2+5.00kg*(3.00m/s)^2-7.00kg*(5m/s)^2}{1120N/m}}\\x=0.25m

We got that the maximum compression is 0.25m.

5 0
3 years ago
An us bomber is flying horizontally at 300 mph at an altitude of 610 m. its target is an iraqi oil tanker crusing 25kph in the s
Pachacha [2.7K]

Answer:

=1419.19 meters.

Explanation:

The time it takes for the shell to drop to the tanker from the height, H =1/2gt²

610m=1/2×9.8×t²

t²=(610m×2)/9.8m/s²

t²=124.49s²

t=11.16 s

Therefore, it takes 11.16 seconds for a free fall from a height of 610m

Range= Initial velocity×time taken to hit the tanker.

R=v₁t

Lets change 300 mph to kph.

=300×1.60934 =482.802 kph

Relative velocity=482.802 kph-25 kph

=457.802 kph

Lets change 11.16 seconds to hours.

=11.16/(3600)

=0.0031 hours.

R=v₁t

=457.802 kph × 0.0031 hours.

=1.41918 km

=1.41919 km × 1000m/km

=1419.19 meters.

3 0
4 years ago
The specific heat of substance X is 200 J/g*C, how much heat is required to change the temperature of 2.0 grams of substance X f
Margarita [4]
C = 200 j/g*C
m = 2g
ΔT = 55 - 35 = 20C

Heat = mcΔT = 2 * 200 * 20 = 8000J = 8kJ
8 0
4 years ago
A light beam is traveling through an unknown substance. When it exits that substance and enters into air, the angle of reflectio
bogdanovich [222]

Answer:

0.79

Explanation:

Using Snell's law, we have that:

n(1) * sin θ1 = n(2) * sinθ2

Where n(1) = refractive index of air = 1.0003

θ1 = angle of incidence

n(2) = refractive index of second substance

θ2 = angle of refraction

The angle of reflection through the unknown substance is the same as the angle of incidence of air. This means that θ1 = 32°

=> 1.0003 * sin32 = n(2) * sin42

n(2) = (1.0003 * sin32) / sin42

n(2) = 0.79

3 0
3 years ago
Read 2 more answers
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