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Eduardwww [97]
3 years ago
14

How often does a Lunar Eclipse occur?

Physics
2 answers:
mash [69]3 years ago
7 0

Answer:

On average, a total lunar eclipse will happen a bit more frequently than twice every three years. Every three and a half years, two total lunar eclipses will happen within the same year. Every 200 years, three total lunar eclipses happen in the same year.

Explanation:

anyanavicka [17]3 years ago
6 0

2-5 a year I believe.

At least two lunar eclipses and as many as five occur every year, although total lunar eclipses are significantly less common. If the date and time of an eclipse is known, the occurrences. On average, a total lunar eclipse will happen a bit more frequently than twice every three years. Every three and a half years, two total lunar eclipses will happen within the same year. Every 200 years, three total lunar eclipses happen in the same year

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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w
Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

=  (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0
3 years ago
A constant horizontal F force began to act on the initially immovable body placed on a horizontal surface. After t time the forc
mel-nik [20]

Answer:

The coefficient of friction is (F/(19.6·m)

Explanation:

The given parameters are;

The force applied to the immovable body = F

The time duration the force acts = t

The time the body spends in motion = 3·t

The acceleration due to gravity, g = 9.8 m/s²

From Newton's second law of motion, we have;

The impulse of the force = F × t = m × Δv₁

Where;

Δv₁ = v₁ - 0 = v₁

The impulse applied by the force of friction, F_f is F_f × (3·t - t) =  F_f × (2·t)

Given that the motion of the object is stopped by the frictional force, we have;

The impulse due to the frictional force = Momentum change = m × Δv₂ = F_f × (2·t)

Where;

Δv₂ = v₂ - 0 = v₂

Given that the velocity, v₂, at the start of the deceleration = The velocity at the point the force ceased to  act, v₁, we have;

m × Δv₂ = F_f × (2·t) = m × Δv₁ = F × t

∴ F_f × (2·t) = F × t

F_f = F × t/(2·t) = F/2

The coefficient of dynamic friction, \mu _k = Frictional force/(The weight of the body) = (F/2)/(9.8 × m)

\mu _k = (F/(19.6·m)

The coefficient of friction, \mu _k = (F/(19.6·m)

5 0
3 years ago
The distance from earth to a NASA satellite traveling through the solar system is 6.0 × 109m. A command is sent to the satellite
alexandr1967 [171]

Answer:

the signal it takes 40 s to get the answer

Explanation:

As the satellite is in space the speed of radiation (waves9 is the speed of light in a vacuum, which is constant 3. 108 m / s, so we can use the uniform motion ratios to find the time.

    v = d / t

    t = d / v

    t = 6.0 109 / 3. 108

    t = 2 101 s

    t = 20 s

This is the time it takes to get the signal from the earth to the satellite and it takes the same time to return, so since they send the signal it takes 40 s to get the answer

5 0
4 years ago
Which of the following better describe planets most comparable to Earth?
wariber [46]

Answer:

??????¿?????????????????????¿

8 0
3 years ago
Read 2 more answers
(a) How can a driver steer a car traveling at constant speed so that the acceleration is zero? (Assume that the road is level. S
PilotLPTM [1.2K]

Explanation:

(a) We know that the acceleration of the car is given by :

a = change in speed / time taken

If the speed of the car is constant in a straight line, the acceleration of the car is zero because there is no change in the speed of the car.

(b) For the driver steer a car traveling at constant speed so that the magnitude of the acceleration remains constant, the driver should drive the car in the circular path. This is because, in circular path the speed of an object remains the same while its velocity changes.

8 0
3 years ago
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