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ira [324]
3 years ago
13

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?​

Physics
1 answer:
stiv31 [10]3 years ago
5 0

Answer:

Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.

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Everybody can you post the lyrics to your favorite song? I'll give brainliest if I know the song or if I like the lyrics, I'll b
andrezito [222]

Answer:

Show and tell

I'm on display for all you (bad word) to see

Show and tell

Harsh words if you don't get a pic with me

Buy and sell (buy and sell me, baby)

Like I'm a product to society

Art don't sell

Unless you (bad word) every authority

Explanation:

4 0
3 years ago
Read 2 more answers
An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1
Svetlanka [38]

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

4 0
3 years ago
e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the
anastassius [24]
The force acting between the particles is

F=k \frac{Q_{1}Q_{2}}{r^2}
Then
Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C




7 0
3 years ago
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iragen [17]
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8 0
2 years ago
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If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?
AleksandrR [38]
Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.
3 0
3 years ago
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