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Rainbow [258]
2 years ago
10

Why aren't there lunar and solar eclipses every month?

Physics
2 answers:
Helga [31]2 years ago
4 0

The reason why lunar and solar eclipses do not occur every month is because; the plane of orbit of the moon around the earth is not same as that of earth around the sun.

<h3>Why don't lunar and solar eclipses occur every month?</h3>

When the Moon passes between Earth and the Sun, there's solar eclipse while when the earth passes between the Sun and the Moon, casting a shadow on the Moon, there's lunar eclipse.

Hence, because the plane of orbit of the moon around the earth is not same as that of earth around the sun, we do not have lunar and solar eclipses every month.

Read more on solar and lunar eclipse;

brainly.com/question/14198324

#SPJ11

Shalnov [3]2 years ago
3 0

The moon's orbital plane is different from that of the earth's orbit around the sun due to which lunar and solar eclipses do not occur every month.

<h3 /><h3>What is a solar eclipse?</h3>

When the moon passes in front of Earth and the sun, it creates a solar eclipse and casts a shadow across the globe.

Only during the new moon phase, when the moon travels squarely between the sun and Earth and casts shadows on its surface, can a solar eclipse occur.

Due to the moon's orbit being five degrees off-center from Earth's orbit around the Sun, lunar eclipses do not happen every month and the Moon often passes above or below the shadow.

Lunar eclipses would happen every month if there were no tilt. Eclipses of the sun and the moon happen equally often.

Hence lunar and solar eclipses aren't coming every month.

To learn more about the solar eclipse refer;

brainly.com/question/17749647

#SPJ1

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A 12,000-N car is raised using a hydraulic lift, which consists of a U-tube with arms of unequal areas, filled with incompressib
victus00 [196]

Answer:

926 N

Explanation:

Metric unit conversion:

R = 18 cm = 0.18 m

r = 5 cm = 0.05 m

The pressure exerted by the F = 12000N car on the wider arm would be ratio of the gravity over area

P = F/A = \frac{F}\pi R^2} = \frac{12000}{2*\pi*0.18^2} = 117892 Pa

The pressure must be the same on the smaller pressure for it to be able to start lifting the car. We can calculate the force f acting on it:

f = Pa = P\pi r^2 = 117892 * \pi * 0.05^2 = 926 N

5 0
3 years ago
Three objects are positioned along the x axis as follows: 4.4 kg at x = + 1.1 m, 3.7 kg at x = +0.80 m, and 2.9 kg at x = +1.6 m
maxonik [38]

Answer:

the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m

Explanation:

Given that

m₁=4.4kg

x₁=+1.1m

m₂=3.7kg

x₂=+0.80m

m₃=2.9kg

x₃=+1.6m

The position of the center of mass is

         Xcm = [m₁x₁ +m₂x₂ +m₃x₃]/(m₁+m₂+m₃)

                = [(4.40kg)(1.1 m)+(3.70 kg)(0.80 m)+(2.90 kg)(1.60 m)]/(4.4 kg + 3.70 kg+2.90 kg)

              = 1.13 m

The position of the center of gravity is 1.13m

Therefore, the distance from the location of the center of gravity to the location of the center o mass for this system is 1.13m

4 0
3 years ago
Calculate the maximum absolute uncertainty for R if:
Radda [10]

Answer:

ΔR = 9 s

Explanation:

To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value

           

The given expression is      R = 2A / B

the uncertainty is                 ΔR = | \frac{dR}{dA} | ΔA + | \frac{ dR}{dB} | ΔB

we look for the derivatives

     \frac{dR}{dA} = 9 / B

     \frac{dR}{dB} = 9A ( - \frac{1}{B^2 } )

we substitute

     ΔR = \frac{9}{B}  ΔA + \frac{9A}{B^2}  ΔB

the values ​​are

     ΔA = 2 s

     ΔB = 3 s

 

     ΔR = \frac{9}{11}   2 + \frac{9 \ 32}{11^2 }  3

     ΔR = 1.636 + 7.14

     ΔR = 8,776 s

the absolute error must be given with a significant figure

     ΔR = 9 s

3 0
3 years ago
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