Answer:
3.4 x 10⁴ m/s
Explanation:
Analyze the circular path of the electron
B = magnetic field = 80 x 10⁻⁶ T
m = mass of an electron = 9.1 x 10⁻³¹ kg
v = speed in the radial direction
r = radius of the circular trajectory = 2 mm = 0.002 m
q = charge of an electron = 1.6 x 10⁻¹⁹ C
For the electron’s circular movement
qBr = mv
(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v
v = 2.8 x 10⁴ m/s
Now, consider the electron's movement in a straight line:
v' = speed in linear motion
x = distance traveled horizontally = 9 mm = 0.009 m
t = duration =
=
= 4.5 x 10⁻⁷ sec
Using the formula
x = v' t
0.009 = v' (4.5 x 10⁻⁷)
v' = 20000 m/s
v' = 2 x 10⁴ m/s
The resultant speed is given by
V = sqrt(v² + v'²)
V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)
v = 3.4 x 10⁴ m/s
The answer is option C. SWOT analysis. The compilation of decisions made by managers to aid the organization in reaching its objectives is referred to as SWOT Analysis.
Here are the options. A. Strategy B. Scenario planning C. SWOT analysis D. Diversification E. Related diversification
The electric flux through the cylindrical surface surrounding the infinite charged wire is given by the formula ∅E = E x 2πrl. To analyze this, we consider an infinitely long straight wire with a uniform linear charge density of λ Cm⁻¹. The electric field at a distance r from this charge can be evaluated using a cylindrical Gaussian surface of radius r and length l, oriented along the wire. Only the curved surface of the cylinder contributes to the total flux since the other surfaces are perpendicular to E.
Answer:
35.79 meters
Explanation:
We have an archer, and there is a target. Denote the distance between them as d.
The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:

Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time
. Thus, we can derive:
.
Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:
.
Using this relationship in the distance formula for sound allows us to write:
.
Substituting the value of d from the first equation yields:
.
Now, after some calculations, we can proceed further:
.
Finally, the value is inserted into the initial equation:


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)
