Answer:
2.83
Explanation:
Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:
(1)
with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):
(2)
Because in the right side of the equation (2) we have only constant quantities, that implies the ratio 
is constant for all the planets orbiting the same sun, so we can said that:
When someone is holding something that has been struck or splashed by lightning, contact damage occurs.
We need additional information concerning lightning and injuries in order to identify the solution.
<h3>What types of injuries are brought on by lightning?</h3>
- Lightning is the name for a natural electrical discharge that occurs quickly and with a dazzling flash.
- It has a tremendous amount of energy.
- Lightning-related injuries can be divided into three categories: direct strikes, side splashes, and contact injuries.
- When someone is struck by lightning directly, they can get direct injury.
- When a current splashes from a neighboring object, it is called a side splash.
- When someone touches a lightning-hit object, contact harm results.
In light of this, we can say that contact injuries happen when a person is holding an object that has been struck by lightning or splashed by it.
Learn more about the lightning and harm here:
brainly.com/question/28055828
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Answer:
Somewhere between the two wires, but closer to the wire carrying λ₂
Explanation:
Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².
Electric Fied due to an electric charge is a vector and its direction is such that if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)
According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires are opposite
In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.
As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.
But for points closer to wire with λ₂ ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance to get equals E and then Ef = 0
Answer:
Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.
Explanation:
For the rod 1 the angular acceleration is
Similarly, for rod 2
Now, the moment of inertia for rod 1 is
,
and the torque acting on it is (about the center of mass)
therefore, the angular acceleration of rod 1 is
Now, for rod 2 the moment of inertia is
and the torque acting is (about the center of mass)
therefore, the angular acceleration 
is
We see here that
therefore
In other words , the initial angular acceleration for rod 1 is greater than for rod 2.
