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expeople1 [14]
3 years ago
10

In the picture of a planet in orbit around its star area a is double that of area b. What is true of the time the planet takes t

o travel A1A2 as compared to B1B2

Physics
1 answer:
aalyn [17]3 years ago
8 0
The Kepler's laws predict the planetary motion, so there are three laws for this, namely:

1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.

2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.
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Water at room temperature is discharged from a pipe at a rate of 1000 gallons per minute (gpm). Express this flow rate in cubic
marshall27 [118]

Answer

given,

discharge rate from pipe = 1000 gallons/minutes

now,

flow rate in  cubic meters per second

1 gallon = 0.00378541 m³

1 min = 60 s

Q = 1000\times \dfrac{0.00378541\ m^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 0.063 m³/s

flow rate in  liters per minute

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flow rate in cubic feet per second

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Q = 1000\times \dfrac{0.133681\ ft^3}{1\ gallon}\times \dfrac{1\ min}{60\ s}

Q = 2.23 ft³/s

4 0
3 years ago
A 53-N force is needed to keep a 50.0-kg box sliding across a flat surface at a constant velocity. What is the coefficient of ki
earnstyle [38]

The weight of the box is (mass) x (gravity) = (50 kg) x (9.8m/s²) = 490 newtons.

If the box is sliding at constant speed, and not speeding up or slowing down,
that means that the horizontal forces on it add up to zero. 

Since you're pushing on it with 53N in <em><u>that</u></em> direction, friction must be pulling
on it with 53N in the <u><em>other</em></u> direction.

 The 53N of friction is (the weight) x (the coefficient of kinetic friction).

                                                  53N  =  (490N) x (coefficient).

Divide each side by  490N :  Coefficient = (53N) / (490N)  =  0.1082 .

Rounded to the nearest hundredth, that's    <em>0.11 </em>.      (choice 'd')


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