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stealth61 [152]
3 years ago
9

What event, which occurs twice a year when the sun crosses Earth's equator, recently

Physics
1 answer:
andrey2020 [161]3 years ago
5 0

Answer:

<u>the autumnal equinox (when day and night are of equal length)</u>

<u>Explanation:</u>

The first event occurs around the 22nd or 23rd in the month of September. What usually happens is that regions around the Northern Hemisphere (which includes all of North America and Europe) experience an equal length of day and night.

While the second event occurs around March 20th or 21st mainly observable in countries found in the Southern Hemisphere.

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When monochromatic light illuminates a grating with 7000 lines per centimeter, its second order maximum is at 62.4°. what is the
zhannawk [14.2K]

When red light illuminates a grating with 7000 lines per centimeter, its second maximum is at 62.4°. What is the wavelength of this light?

ans: 633nm

6 0
3 years ago
Read 2 more answers
Lucille is finding it difficult to play soccer after school. Her doctor thinks that her cells might not be getting enough oxygen
lana66690 [7]

Answer:

See the explanation below.

Explanation:

Circulation of blood and oxygen is possible in body when circulatory system work along with respiratory system. Through tranche air moves in and out from lungs, whereas, through pulmonary arteries and veins (both connected to heart) blood moves in and out from the lungs. As Lucille is facing problem in his respiratory and circulatory system hence, it is difficult for him to play soccer because under normal circumstances when there is increase in physical activity then muscle cell respire more as compare to when the body is on rest. So, with increase of physical activity there is also increase in the rate of breathing which result in more absorption of oxygen and more removal of carbon dioxide but if there is problem in respiratory and circulatory system, for example, infection in throat due to pollution,etc. then this normal process of breathing gets affected which sometime may prove fatal to the person.

4 0
2 years ago
Does water pressure depend on the total amount of water present? <br><br> (Will pick brainliest)
yaroslaw [1]

Answer:

No, because pressure is determined by force and the area over which that force acts.

Explanation:

3 0
3 years ago
During their physics field trip to the amusement park, Leslie and Maria took a ride on the Whirligig. The Whirligig ride consist
grandymaker [24]

Answer:

a) frequency = 0.1724 Hz

b) Period = 5.8 sec

c) speed = 7.04 m/s

d) acceleration = 7.62 m/s²

Explanation:

Given that;

radius = 6.5m

time period = 5.8 sec every circle

a)  the frequency

frequency is the number of rotation in unit time

frequency = 1 / time period = 1/5.8

frequency = 0.1724 Hz

b)  the period

period is time taken in one rotation

period = total time / rotation = 5.8 / 1

Period = 5.8 sec

c)  the speed

speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8

speed = 7.04 m/s

d) acceleration

To find the acceleration we take the linear velocity squared divided by the radius of the circle.

so

acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5

acceleration = 7.62 m/s²

6 0
3 years ago
Consider an electron with charge −e and mass m orbiting in a circle around a hydrogen nucleus (a single proton) with charge +e.
alexandr1967 [171]

Answer:

v=\sqrt{k\frac{e^2}{m_e r}}, 2.18\cdot 10^6 m/s

Explanation:

The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k is the Coulomb constant

e is the magnitude of the charge of the electron

e is the magnitude of the charge of the proton in the nucleus

r is the distance between the electron and the nucleus

v is the speed of the electron

m_e is the mass of the electron

Solving for v, we find

v=\sqrt{k\frac{e^2}{m_e r}}

Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately

r=5.3\cdot 10^{-11}m

while the electron mass is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

Substituting into the formula, we find

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

7 0
3 years ago
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