Complete Question
An aluminum "12 gauge" wire measures a diameter of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field E in the wire varies over time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is recorded in seconds.
At time 5 seconds, I = 1.2 A.
We need to find the charge Q traveling through a cross-section of the conductor from time 0 to time 5 seconds.
Answer:
The charge is 
Explanation:
The question indicates that
The wire’s diameter is 
The radius of the wire is 
Aluminum's resistivity is 
The electric field variation is described as
The charge is effectively given by the equation
Where A is the area expressed as
Thus,
Therefore
By 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.)
The question states that t = 5 seconds
![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)
The star is moving away from our planet. To elaborate on the Doppler shift: This phenomenon, related to the Doppler effect, is the variation in the perceived frequency or wavelength (color) of a wave when the source of the waves and the observer are in motion relative to one another. Consequently, it can be inferred that as an object recedes, it exhibits more redshift in its spectrum. For instance, when a star moves away, its spectral lines shift towards the red end of the spectrum, whereas if it approaches Earth, the spectral lines move towards blue. Given that the peak wavelength is roughly 650 nm—which is associated with red—it can be concluded that the star is indeed moving away from Earth.
Answer:
Arrangement A facilitates a greater rate of heat transfer
Explanation:
The rate of heat conduction between two materials differing in temperature adheres to the formula
This equation demonstrates that
- An enlarged equivalent cross-sectional area between the two materials will enhance the rate of heat transfer.
- A reduction in the length of the medium will also amplify the rate of heat transfer
Arrangement A features a shorter transfer medium and a larger equivalent cross-sectional area, since the two rods are set up in a parallel arrangement.
Conversely, Arrangement B has a longer transfer medium and a smaller cross-sectional area, as the two rods are placed end to end
Answer:
4.05 m/s
Explanation:
We will express the varying velocities as vectors.
Newton moves northward at 3.90 m/s from Daniel's stationary position.
V_n = 3.9 j
Assuming Pauli runs relative to Daniel at velocity X.
The relative velocity of Newton as seen by Pauli will be
3.9 j - X
Given that
the relative velocity of Newton with respect to moving Pauli = 1.1 i (1.1 towards the east).
Thus,
3.9 j - X = 1.1 i
X = -1.1 i + 3.9 j.
Magnitude of X
X² = 1.1² + 3.9²
X = 4.05 m/s
Therefore, Pauli runs relative to Daniel at 4.05 m/s.
The direction will be west of north at an angle θ,
Tan θ = 1.1 / 3.9
