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Flura [38]
3 years ago
11

You see a total eclipse of the sun. Three Saros cycles later, you find yourself at the same location.

Physics
1 answer:
alukav5142 [94]3 years ago
5 0

Answer:

The correct answer is B. You would not have to travel far to see another total eclipse  .

Explanation:

it is said that the way to predict or know beforehand if there is a lunar or solar eclipse is just to know the what is called the saros which are periods of exactly 223 synodic months and the eclipses are repeated or occurred in the same location every 3 Saros cycles.

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What is the correct answer?
pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

    -  x^3-4x^2             <-- (x-4)(x^2)

_________________

              6x^2-16x-32

           -  6x^2-24x     <-- (x-4)(6x)

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                          8x-32

                       -  8x-32 <- (x-4)(8)

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                                 0 | x^2+6x+8

This means the answer is B) x^2+6x+8

             

3 0
3 years ago
How to make a paper airplane curve down and fly upside down?
Ainat [17]
Throw it sideways and try to make it spin around but it needs to be thrown high up then it should kinda glide down
8 0
3 years ago
Read 2 more answers
Can someone help asap plz don’t understand how to do this
shtirl [24]
Homie I don’t know either‍♀️Drop out of schl ig ;-;
3 0
3 years ago
What happens to the mass and to the weight of the foam rubber when it is compressed?​
Orlov [11]

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423523523523

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4 0
2 years ago
Read 2 more answers
A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
3241004551 [841]

Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

The initial total mass of the rocket is:

m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

7 0
3 years ago
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