Answer:
Explanation:
Given that,
The compression in the spring, x = 0.0647 m
Speed of the object, v = 2.08 m/s
To find,
Angular frequency of the object.
Solution,
We know that the elation between the amplitude and the angular frequency in SHM is given by :
A is the amplitude
In case of spring the compression in the spring is equal to its amplitude
So, the angular frequency of the spring is 32.14 rad/s.
Answer:
In a velocity selector, there are two forces namely;
» Electric field Intensity
» Magnetic field density
<u>Relationship</u><u>:</u>
E is the electric field intensity
B is the magnetic flux density
Answer:
electrons
Explanation:
An electric current is said to exist when there is a net flow of electric charge through a region. In electric circuits this charge is often carried by electrons moving through a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionized gas (plasma).
Answer:
Cosmic ray's frame of reference: 99,875 years
Stationary frame of reference: 501,891 years
Explanation:
First of all, we convert the distance from parsec into metres:
The speed of the cosmic ray is
where
is the speed of light. Substituting,
And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:
Converting into years,
Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:
And substituting v = 0.98c, we find:
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
