When the mass of the sun is larger, Earth moves around the sun at a faster or slower pace.

1 answer:

4 0

-- The mass of the sun never increases.

-- It does decrease, but not nearly enough to have any noticeable
effect on the orbital motion of the Earth, or any other planet.

-- When Earth is closer to the sun, it moves faster in its orbit.

-- When Earth is farther from the sun, it moves slower in its orbit.

-- The result is that the line from the sun to the Earth always covers
the same amount of area in the same length of time.

-- Johannes Kepler noticed this, and it's his Second Law of planetary motion.

-- Newton showed that if his equations for gravity and motion are correct,
then planets MUST behave this way.

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Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200

artcher [175]

We have all the charges for q1, q2, and q3.
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 | / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2
r^2 x F(3on1) = k |q1 q3|
r = sqrt of k |q1 q3| / F(3on1)
= .144 m (distance between q1 and q3)
0 - .144m

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

6 0

3 years ago

Read 2 more answers

What is the thermal energy of an object?

Vedmedyk [2.9K]

The total kinetic and potential energies

3 0

3 years ago

Telephone signals are often transmitted over long distances by microwaves. What is the frequency of microwave radiation with a w

amm1812

Answer:

1) f= 8.6 GHz

2) t= 0.2 ms

Explanation:

Since microwaves are electromagnetic waves, they move at the same speed as the light in vacuum, i.e. 3*10⁸ m/s.
There exists a fixed relationship between the frequency (f) , the wavelength (λ) and the propagation speed in any wave, as follows:

$v = \lambda * f (1)$

Replacing by the givens, and solving for f, we get:

$f =\frac{c}{\lambda} =\frac{3e8m/s}{0.035m} = 8.57e9 Hz (2)$

⇒ f = 8.6 Ghz (with two significative figures)

Assuming that the microwaves travel at a constant speed in a straight line (behaving like rays) , we can apply the definition of average velocity, as follows:

$v =\frac{d}{t} (3)$