Answer:
velocity = 472 m/s
velocity = 52.4 m/s
Explanation:
given data
steady rate = 0.750 m³/s
diameter = 4.50 cm
solution
we use here flow rate formula that is
flow rate = Area × velocity .............1
0.750 = 
× (4.50×
)² × velocity
solve it we get
velocity = 472 m/s
and
when it 3 time diameter
put valuer in equation 1
0.750 = × 3 × (4.50×)² × velocity
velocity = 52.4 m/s
Magnets facing the same way <span />
The frequency, f, of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz (Hz), since one hertz is equal to one wave per second.
Neutrons are neutral<span> and </span>do<span> not </span>have<span> any </span>charge<span> at all. Protons carry a positive </span>charge<span>, and electrons carry the negative </span><span>charge.</span>
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
