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Mice21 [21]
4 years ago
8

During a lunar eclipse the mood is not completely dark, but it is often a deep red in color. Explain this in terms of the refrac

tion of all the sunsets and sunrises around the world.
Physics
2 answers:
Anna35 [415]4 years ago
4 0
The Moon does not completely disappear as it passes through the umbra because of the refraction of sunlight by the Earth's atmosphere into the shadow cone; if the Earth had no atmosphere, the Moon would be completely dark during an eclipse. The red colouring arises because sunlight reaching the Moon must pass through a long and dense layer of the Earth's atmosphere, where it is scattered. Shorter wavelengths are more likely to be scattered by the small particles, and so by the time the light has passed through the atmosphere, the longer wavelengths dominate. This resulting light we perceive as red. This is the same effect that causes sunsets and sunrises to turn the sky a reddish color; an alternative way of considering the problem is to realise that, as viewed from the Moon, the Sun would appear to be setting (or rising) behind the Earth.
Dmitry [639]4 years ago
4 0
<span>The answer to this question is the same as the answer to why is the sky blue? Refraction or the suns rays causes blue light to pass straight down to the earth. Red light does not come straight down. that is why sunset and sunrise is red. 

This red light hits the moon during eclipses that are near blackouts of the moon</span>
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In a mass spectrometer used for commercial purposes, uranium ions of mass 3.76 X 10^(-25) kg and charge 3.5 X 10^(-19) C are sep
Flauer [41]

Answer:

a. 0.394 T b. 0.255 A c. 1.309 × 10⁸ J

Explanation:

Here is the complete question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 109 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.31 m. After traveling through 180° and passing through a slit of width 0.752 mm and height 0.991 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 1.12 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.31 h.

Solution

a. The magnitude of the (perpendicular) magnetic field in the separator

The kinetic energy of the uranium ions = electric potential energy

¹/₂mv² = qV

v = √(2qV/m) where v = speed of uranium ions, q = uranium ion charge = 3.2 × 10⁻¹⁹ C , m = mass of uranium ions = 3.92 × 10⁻²⁵ kg and V = 109 kV = 1.09 × 10⁵ V

v = √(2qV/m) = √(2 × 3.2 × 10⁻¹⁹ C × 1.09 × 10⁵ V/3.92 × 10⁻²⁵ kg)

v = 4.22 × 10⁵ m/s

Also, the magnetic force on the uranium ions equals the centripetal force when it passes through the magnetic field.

Bqv = mv²/r

B = mv/rq   where B = magnetic field strength and r = radius of circle = 1.31 m

B = m(√(2qV/m))/rq

B = √(2mV/q)/r

B = √(2 × 3.92 × 10⁻²⁵ kg × 1.09 × 10⁵ V/3.2 × 10⁻¹⁹ C)/1.31 m

B = √0.26705/1.31

B = 0.394 T

(b) the current (in A) of the desired ions in the machine

Since a mass m of 3.92 × 10⁻²⁵ kg of uranium ions carries a charge q of 3.2 × 10⁻¹⁹ C, then 1.12 mg per hour = 1.12 × 10⁻³ kg/h. In 1.31 h, our mass is M = 1.12 × 10⁻³ kg/h × 1.31 h = 1.47 × 10⁻³ kg carries a charge of Q of

m/q = M/Q

Q = Mq/m

Q = 1.47 × 10⁻³ kg × 3.2 × 10⁻¹⁹ C/3.92 × 10⁻²⁵ kg

Q = 1200 C

The current i = Q/t where t = time = 1.31 h = 1.31 × 60 × 60 s = 4716 s

i = 1200/4716

i = 0.2545 A

i ≅ 0.255 A

(c) the thermal energy (in J) produced in the cup in 1.31 h.

The thermal energy produced in the cup equals the kinetic energy lost by the uranium ions hitting the cup in 1.31 h.

E = ¹/₂Mv² = ¹/₂ × 1.47 × 10⁻³ kg × (4.22 × 10⁵ m/s)²

E = 1.309 × 10⁸ J

3 0
4 years ago
Which of the following is a vector quantity?
mestny [16]
A is the answer  of course
5 0
3 years ago
Read 2 more answers
A child in a boat throws a 5.3 kg package out horizontally with a speed of 10.0m/s. calculate the velocity of the boat immediate
seropon [69]
<span>(Momentum is Zero) 0 = (26.0 kg + 55.0 kg)v + (5.40 kg)(+10.0 m/s)
</span>velocity = -.667 m/s
5 0
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A single roller-coaster car is moving with speed v0 on the first section of track when it descends a 4.7-m-deep valley, then cli
yan [13]

Explanation:

From the curve given in the attachment,

a) minimum speed v0 required if the car is to travel beyond the top of the hill

climb to the top = 5.4 m

Therefore, V_0=\sqrt{2gh}

V_0=\sqrt{2(9.81)(5.4)} =  10.29 m/s

b) No, we cannot affect this speed by changing the depth of the valley to make the coaster pick up more speed at the bottom as this speed does not depend upon depth of the valley.

5 0
3 years ago
A cat with a mass of 5.00 kg pushes on a 25.0 kg desk with a force of 50.0N to jump off. What is the force on the desk?
trasher [3.6K]

Answer:

i dont know

Explanation:

Sorry and good luck

3 0
3 years ago
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