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8090 [49]
3 years ago
11

What would you say to a friend who made this statement, “The visible-light spectrum of the Sun shows weak hydrogen lines and str

ong calcium lines. The Sun must therefore contain more calcium than hydrogen.”?
Physics
1 answer:
Oliga [24]3 years ago
8 0

Explanation:

spectral lines or signatures of elements depend on temperature, the temperature of the sun is about 5800 K.

at this temperature most calcium atoms are excited to higher energy states than hydrogen atoms and this means that calcium atoms are gonna have more signatures than the atoms of hydrogen.

the statement that the sun shows weak hyrogen lines and strong calcium line is wrong because at the sun's temperature most of the hydrogen atoms are in lower energy states while calcium atoms are in higher energy states hence calcium has more or ''strong'' lines than hydrogen.

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Where on the Earth would you have to be to see the North Celestial Pole exactly halfway between your horizon and zenith?
mario62 [17]

Answer:

The point straight overhead on the celestial sphere for any observer is called the zenith and is always 90 degrees from the horizon. The arc that goes through the north point on the horizon, zenith, and south point on the horizon is called the meridian.

From any location on Earth you see only half of the celestial sphere, the half above the horizon.

If you stood at the North Pole of Earth, for example, you would see the north celestial pole overhead, at your zenith. The celestial equator, 90° from the celestial poles, would lie along your horizon.

4 0
3 years ago
what will be the acceleration due to gravity at up planet whose mass is 8 times the mass of the earth and whose radius is twice
Alexxandr [17]
Hope it cleared your doubt.

3 0
3 years ago
There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co
mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

q_1=q_2=2.06\mu C=2.06\times 10^{-6} C

q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C

Distance,d=0.41 m

The magnitude of the net electrostatic force experienced by any charge at point 4

Net force,F_{net}=\sqrt{F^2_1+F^2_3+2F_1F_3cos90^{\circ}}-F_2

F_1=F_3=F

F_{net}=\sqrt{F^2+F^2+0}-F_2

F_{net}=\sqrt 2F-F_2

F=\frac{kq^2}{d^2}

F_2=\frac{Kq^2}{2d^2}

F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})

Where k=9\times 10^9

F_{net}=\frac{9\times 10^9\times (2.06\times 10^{-6})^2}{(0.41)^2}(\sqrt 2-\frac{1}{2})

F_{net}=0.208 N

3 0
3 years ago
Electromagnetism consists of what two processes?
kherson [118]

Answer: it’s A and B

Explanation: everyone else on this post was giving you the wrong answer.

5 0
3 years ago
Read 2 more answers
A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?
Anestetic [448]

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2  

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

8 0
3 years ago
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