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swat32
3 years ago
9

A total solar eclipse is a rare event. Although they occur somewhere on earth every 18 months on average, it is estimated that t

hey recur at any given place only once every 370 years. Why are reoccurring solar eclipses so rare?
Physics
2 answers:
WITCHER [35]3 years ago
6 0

Answer:

D

Explanation:

The Moon passes slightly above or below the line between the Sun and the Earth because of its tilted orbit. I also had it for my USA Test Prep

Setler79 [48]3 years ago
3 0
Because the tip of the moon's shadow ... the area of "totality" ... is never more than a couple hundred miles across, It never covers a single place for more than 7 minutes, and can never stay on the Earth's surface for more than a few hours altogether during one eclipse.

If you're not inside that small area, you don't see a total eclipse.
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Ne4ueva [31]
Moving water, wind, gravity, and ice are the five natural agents of erosion
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Thylakoids contain chlorophyl that absorb solar energy
steposvetlana [31]

Answer:

True

True statement:

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3 0
3 years ago
The greater the mass of an object
Morgarella [4.7K]
The answer is option C.
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7 0
3 years ago
An iron bal of mass 20kg is rolling on a flat surface. On applying force, the velocity change from 17ms to 27m/s in 5s. Calculat
pentagon [3]

Answer:

40 N

Explanation:

We first need to calculate the acceleration of the tron ball.

Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.

So, a = (v - u)/t

= (27 m/s - 17 m/s)/5 s

= 10 m/s ÷ 5 s

= 2 m/s²

We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²

So, F = ma

= 20 kg × 2 m/s²

= 40 kgm/s²

= 40 N

So, the magnitude of the force on the iron ball is 40 N.

4 0
3 years ago
a car accelerates uniformly from rest to a speed of 65 km/h (18 m/s) in 12s. Find the distance the car travels during this time?
Vinil7 [7]

The car's speed was zero at the beginning of the 12 seconds,
and 18 m/s at the end of it.  Since the acceleration was 'uniform'
during that time, the car's average speed was (1/2)(0 + 18) = 9 m/s.

12 seconds at an average speed of 9 m/s  ==>  (12 x 9) = 108 meters .

==========================================

That's the way I like to brain it out.  If you prefer to use the formula,
the first problem you run into is:  You need to remember the formula !

The formula is        D = 1/2 a T²

                   Distance = (1/2 acceleration) x (time in seconds)²

             Acceleration = (change in speed) / (time for the change)
                                  =        (18 m/s)            /        (12 sec)
                                  =                      1.5 m/s² .

                  Distance  =  (1/2 x 1.5 m/s²) x (12 sec)²
                                  =       (0.75 m/s²)  x  (144 sec²)  =  108 meters .

5 0
3 years ago
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