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)
Answer:
a =( -0.32 i ^ - 2,697 j ^) m/s²
Explanation:
This problem is an exercise of movement in two dimensions, the best way to solve it is to decompose the terms and work each axis independently.
Break down the speeds in two moments
initial
v₀ₓ = v₀ cos θ
v₀ₓ = 5.25 cos 35.5
v₀ₓ = 4.27 m / s
= v₀ sin θ
= 5.25 sin35.5
= 3.05 m / s
Final
vₓ = 6.03 cos (-56.7)
vₓ = 3.31 m / s
= v₀ sin θ
= 6.03 sin (-56.7)
= -5.04 m / s
Having the speeds and the time, we can use the definition of average acceleration that is the change of speed in the time order
a = (
- v₀) /t
aₓ = (3.31 -4.27)/3
aₓ = -0.32 m/s²
= (-5.04-3.05)/3
= -2.697 m/s²
Answer:
a) 
b) 
c) 
d) 
e) 
&
f) 
Explanation:
From the question we are told that:
Stretch Length 
Mass 
Total stretch length
a)
Generally the equation for Force F on the spring is mathematically given by
b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by
Where
A=Amplitude
And
Therefore
c)
Generally the equation for Max Acceleration of Mass on the spring is mathematically given by
d)
Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by
e)
Generally the equation for the period T is mathematically given by
Generally the equation for the Frequency is mathematically given by
f)
Generally the Equation of time-dependent vertical position of the mass is mathematically given by
Where
'= signify order of differentiation
Nothing happens to the mirror.
However, if the ray is within some suitable range of
wavelengths, the ray is reflected from the mirror's surface.
