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s344n2d4d5 [400]
3 years ago
5

Explain Kepler's second law in detail with figures.

Physics
1 answer:
Jet001 [13]3 years ago
5 0

Answer:

Kepler's second law of planetary motion describes the speed of a planet traveling in an elliptical orbit around the sun. It states that a line between the sun and the planet sweeps equal areas in equal times. Thus, the speed of the planet increases as it nears the sun and decreases as it recedes from the sun.

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An electric field with a magnitude of 6.0 × 104 N/C is directed parallel to the positive y axis. A particle with a charge q = 4.
jekas [21]

Answer:

Electric force, F=2.88\times 10^{-14}\ N

Explanation:

It is given that,

Magnitude of electric field, E=6\times 10^4\ N/C

Charge, q=4.8\times 10^{-19}\ C

The electric field is directed parallel to the positive y axis. We need to find the  force on the charge particle. It is given by :

F=q\times E

F=4.8\times 10^{-19}\ C\times 6\times 10^4\ N/C

F=2.88\times 10^{-14}\ N

So, the electric force on the charge is 2.88\times 10^{-14}\ N. Hence, this is the required solution.

5 0
4 years ago
Differentiate betwwen speed and velocity
Elanso [62]
Speed is the speed that object is moving on while velocity is the direction and movement speed you can make it but velocity no
4 0
3 years ago
Dana has a sports medal suspended by a long ribbon from her rearview mirror. As she accelerates onto the highway, she notices th
ivann1987 [24]

Answer:

2.26m/s^2

Explanation:

Suppose the tension in the ribbon is T.

The upwards component of the tension = Tcos13

The horizontal component of the tension = Tsin13

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In a vertical direction, mg is balanced by Tcos13, so mg=Tcos13, so T  = mg/cos13. The horizontal component of T is the force causing m to accelerate.  

F = ma\\T\sin13^{\circ} = ma

Substituting T= mg/cos13 gives:

\frac{mg}{cos13^{\circ}}sin13^{\circ} = ma

m cancels and you have

g\tan13 = a\\a=9.81\tan13^{\circ} \approx 2.26m/s^2

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Answer:

10000

Explanation:

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