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2 years ago

Kepler's laws, satellites motion and weightlessness

Physics

1 answer:

Anon25 [30]2 years ago

3 0

Answer:

First Kepler law states that Each planet describes an ellipsoidal motion about the sun as its single focus.

Second Kepler law states that An imaginary line joining a planet to the Sun sweeps out equal areas in equal time intervals.

Third Kepler law states that The squares of period of revolution of the planet around the sun are proportional to the cubes of mean distance between the planet and the sun.

Weightlessness is the condition where the body has zero gravity ( its acceleration is equal to the acceleration due to gravity )

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Can somebody re-answer the true and false questions!!!

stiv31 [10]

Answer:

I think no 3 is false

and 4 is true

and the the ones you did are correct

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3 years ago

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Some birds migrate 20,000 miles. If 1 mile equals 1.6 kilometers, calculate the distance these birds fly in kilometers.

faust18 [17]

Set up a proportion

1 mile/1.6 km = 20,000miles/x

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Answer:

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2 years ago

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Two power lines run parallel for a distance of 222 m and are separated by a distance of 40.0 cm. if the current in each of the t

earnstyle [38]

1) Magnitude of the force:

The magnetic field generated by a current-carrying wire is
$B= \frac{\mu_0I}{2 \pi r}$
where
$\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance at which the field is calculated

Using I=135 A, the current flowing in each wire, we can calculate the magnetic field generated by each wire at distance
$r=40.0 cm=0.40 m$ ,
which is the distance at which the other wire is located:
$B= \frac{\mu_0 I}{2 \pi r}= \frac{(4 \pi \cdot 10^{-7} N/A^2)(135 A) }{2 \pi (0.40 m)}=6.75 \cdot 10^{-5} T$

Then we can calculate the magnitude of the force exerted on each wire by this magnetic field, which is given by:
$F=ILB=(135 A)(222 m)(6.75 \cdot 10^{-5}T)=2.03 N$

2) direction of the force:
The two currents run in opposite direction: this means that the force between them is repulsive. This can be determined by using the right hand rule. Let's apply it to one of the two wires, assuming they are in the horizontal plane, and assuming that the current in the wire on the right is directed northwards:
- the magnetic field produced by the wire on the left at the location of the wire on the right is directed upward (the thumb of the right hand is directed as the current, due south, and the other fingers give the direction of the magnetic field, upward)

Now let's apply the right-hand rule to the wire on the right:
- index finger: current --> northward
- middle finger: magnetic field --> upward
- thumb: force --> due east --> so the force is repulsive

A similar procedure can be used on the wire on the left, finding that the force exerted on it is directed westwards, so the force between the two wires is repulsive.

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2 years ago

3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ

WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

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2 years ago