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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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The engine of a 550kg car is producing a net force of 11,000N. What is its acceleration?
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Answer:

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Explanation:

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3 years ago
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How does the circuit change when the wire is added? a closed circuit occurs and makes all bulbs turn off. an open circuit occurs
Vitek1552 [10]

The circuit change when a wire is added is, an open circuit occurs and makes all bulbs turn off.

<h3>What is a closed circuit?</h3>

A closed circuit is a type of circuit connection in which the wire connection is complete and current flow occurs, turning the light bulbs on in the process.

<h3>What is an open circuit?</h3>

An open circuit is a type of circuit connection in which the wire connection is incomplete and current cannot flow, turning off the light bulbs.

Thus, the circuit change when the wire is added is, an open circuit occurs and makes all bulbs turn off.

Learn more about open circuit here: brainly.com/question/20351910

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3 0
2 years ago
Find an expression for the minimum frictional coefficient needed to keep a car with speed v on a banked turn of radius R designe
solniwko [45]
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr 
</span><span>Θ = arctan(v0² / gr) </span>

<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>

<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
4 0
2 years ago
An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.3
jekas [21]

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0  N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C

4 0
2 years ago
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Please help me ejnfmwkefwefm i dont wanna fail
Rasek [7]

Answer:B

Explanation:

6 0
3 years ago
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