Answer:
1.429*10^-5 m
Explanation:
From the question, we are given that
Diameter of the cable, d = 3 cm = 0.03 m
Force on the cable, F = 2 kN
Young Modulus, Y = 2*10^11 Pa
Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²
The fractional length = Δl/l
Δl/l = F/AY
Δl/l = 2000 / 0.0007 * 2*10^11
Δl/l = 2000 / 1.4*10^8
Δl/l = 1.429*10^-5 m
Therefore, the fractional length is 1.429*10^-5 m long
Answer: A wavelength is how long ONE wave is. First divide the distance of the whole diagram by the number of waves. you will get 2m. This is the answer.
Answer:
Explanation:
Let the potential difference between the middle point and one of the plate be ΔV .
electric potential energy will be lost and it will be converted into kinetic energy .
Electrical potential energy lost = Vq , where q is charge on charge particle .
For proton
ΔV× q = 1/2 M V² ( kinetic energy of proton )
where M is mass and V be final velocity of proton .
For electron
ΔV× q = 1/2 m v² ( kinetic energy of electron )
where m is mass and v be final velocity of electron . Charges on proton and electron are same in magnitude .
As LHS of both the equation are same , RHS will also be same . That means the kinetic energy of both proton and electron will be same
1/2 M V² = 1/2 m v²
(V / v )² = ( m / M )
(V / v ) = √ ( m / M )
In other words , their velocities are inversely proportional to square root of their masses .
Assuming north as positive direction, the initial and final velocities of the ball are:

(with negative sign since it is due south)

the time taken is

, so the average acceleration of the ball is given by

And the positive sign tells us the direction of the acceleration is north.
Complete Question
An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.
I = 1.2 A at time 5 secs.
Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.
Answer:
The charge is 
Explanation:
From the question we are told that
The diameter of the wire is 
The radius of the wire is 
The resistivity of aluminum is 
The electric field change is mathematically defied as

Generally the charge is mathematically represented as

Where A is the area which is mathematically represented as

So

Therefore

substituting values
![Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5Cint%5Climits%5E%7Bt%7D_%7B0%7D%20%7B%20%5B%200.0004t%5E2%20-%200.0001t%20%2B0.0004%5D%20%7D%20%5C%2C%20dt)
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%20t%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
From the question we are told that t = 5 sec
![Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004t%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20t%5E2%7D%7B2%7D%20%2B0.0004t%5D%20%7D%20%20%5Cleft%20%7C%205%7D%20%5Catop%20%7B0%7D%7D%20%5Cright.)
![Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }](https://tex.z-dn.net/?f=Q%20%3D%20120%20%5B%20%5Cfrac%7B0.0004%285%29%5E3%20%7D%7B3%7D%20-%20%5Cfrac%7B0.0001%20%285%29%5E2%7D%7B2%7D%20%2B0.0004%285%29%5D%20%7D)
