The magnitude and direction of the electric field in the wire are mathematically given as
![L &=[(v / L) v / m] \hat{i}](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D)
<h3>What is the magnitude and direction of the electric field in the wire?</h3>
Generally, the equation for is mathematically given as
A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides
E d 
![\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%26-E%20%5Cint_0%5EL%20d%20x%3D%5Cint_v%5E0%20d%20v%20%5C%5C%5Ctherefore%20E%20%5Ccdot%20L%20%26%3Dv%20%5C%5CL%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D%20%5Chat%7Bi%7D%5Cend%7Baligned%7D)
In conclusion, the magnitude and direction of the electric field in the wire are given as
![L &=[(v / L) v / m]](https://tex.z-dn.net/?f=L%20%26%3D%5B%28v%20%2F%20L%29%20v%20%2F%20m%5D)
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Answer:
hellooooo :) ur ans is 33.5 m/s
At time t, the displacement is h/2:
Δy = v₀ t + ½ at²
h/2 = 0 + ½ gt²
h = gt²
At time t+1, the displacement is h.
Δy = v₀ t + ½ at²
h = 0 + ½ g (t + 1)²
h = ½ g (t + 1)²
Set equal and solve for t:
gt² = ½ g (t + 1)²
2t² = (t + 1)²
2t² = t² + 2t + 1
t² − 2t = 1
t² − 2t + 1 = 2
(t − 1)² = 2
t − 1 = ±√2
t = 1 ± √2
Since t > 0, t = 1 + √2. So t+1 = 2 + √2.
At that time, the speed is:
v = at + v₀
v = g (2 + √2) + 0
v = g (2 + √2)
If g = 9.8 m/s², v = 33.5 m/s.
Answer:
The liquid turns to a gas.
Explanation:
If a liquid is heated the particles are given more energy and move faster and faster expanding the liquid. Particles in the middle of the liquid form bubbles of gas in the liquid.
Answer:
sound water because sound is the most important electronic wave for water
Answer:
H = 54.37
Explanation:
given,
lead ball attached at = 1.70 m
rate of revolution = 3 revolution/sec
height above the ground = 2 m

circumference of the circle = 2 π r
= 2 x π x 1.7
= 10.68 m

v = 32.04 m/s
using conservation of energy




H = 54.37
the maximum height reached by the ball is equal to H = 54.37