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charle [14.2K]
2 years ago
11

How do you know that light travels faster than

Physics
1 answer:
Dahasolnce [82]2 years ago
8 0

Answer:

Explanation:

Answer

How about a thunder storm. I don't know if you live in a city or out in the sticks as I do. Lighting is very obvious and it is not a good idea to be out when experiencing a thunderstorm, especially in an open field. You might be the only thing around that will cause the lightning to be connected to the ground.#  Seconds later, you will hear the thunder which is quite harmless. If you live in the city, observing this is not quite so easy. There are all kinds of buildings around and some of them may block your view.

#The great golfer, Lee Travino, was "hit" by lightening storm twice while playing in tournaments.

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Why is earth a magnet?
oksian1 [2.3K]
I hope this helps. it was right when i took the test.

7 0
3 years ago
Ok so I am taking a test I need help please
Aloiza [94]

Answer:

i know its definetly either clockwise or counter clockwise

Explanation:

6 0
3 years ago
a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance
aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b)  Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

  = \frac{KQ}{r^{2} } for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b)  Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{3.10^{2} }

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = \frac{KQ}{r^{2} }

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - \frac{9\times10^{9}\times5.50\times10^{-6}  }{8^{2} }

E = - 0.77 x 10³ V/m

8 0
3 years ago
A baseball has a mass of 0.145 kilograms, and a bowling ball has a mass of 6.8 kilograms. What is the gravitational force betwee
Olegator [25]
Quite low

Gravitation force = \frac{6.67* 10^{-11}*0.145*6.8 }{ 0.5^{2} } = 26.3 * 10^{-11} = 2.63 *10^{-10} N
6 0
3 years ago
Read 2 more answers
If a fish experiences an a of 9.8m/s2 over a period of 1.84s, what was the 4v?
777dan777 [17]

Answer:

It’s 18.0 m/s

Explanation:

Use acceleration formula then plug in 9.8 and 1.84s

6 0
3 years ago
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