Response:
The speed at which the distance from the helicopter to you is changing (in ft/s) after 5 seconds is 
ft/ sec
Clarification:
Provided:
h(t) = 25 ft/sec
x(t) = 10 ft/ sec
h(5) = 25 ft/sec. 5 = 125 ft
x(5) = 10 ft/sec. 5 = 50 ft
At this point, we can determine the distance between the individual and the helicopter utilizing the Pythagorean theorem
Now, let's calculate the derivative of distance in relation to time
By plugging in the values for h(t) and x(t) and simplifying, we arrive at,
= 
= ft / sec
Answer:
10000 V
0.00225988700565 m²
Explanation:
E = Electric field = 
d = Distance = 2.5 mm
Q = Charge = 80 nC
= Permittivity of free space = 
The potential difference is calculated as
The potential difference across the plates amounts to 10000 V
Area is determined by
The area of each plate measures 0.00225988700565 m²
Capacitance is determined by
The capacitance is
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)
<span>E = h x f </span>
<span>Thus: </span>
<span>f = E / h </span>
<span>f = 4.41•10^-19 / 6.62•10^-34 </span>
<span>f = 6.66•10^14 Hz (s^-1) </span>
<span>b/ What is the wavelength of this light? </span>
<span>------------------------------ </span>
<span>λ = c / f </span>
<span>λ = 3•10^8 / 6.66•10^14 </span>
<span>λ = 4.50•10^-7 m </span>
Response:
(a) 104 N
(b) 52 N
Clarification:
Provided Information
Incline angle of the ramp: 20°
F forms a 30° angle with the ramp
The parallel component of F along the ramp is Fx = 90 N.
The perpendicular component of F is Fy.
(a)
Consider the +x direction pointing up the slope, and the +y direction perpendicular to the ramp's surface.
Using the Pythagorean theorem, decompose F into its x-component:
Fx=Fcos30°
To find F:
F= Fx/cos30°
Insert the value for Fx based on the given info:
Fx=90 N/cos30°
=104 N
(b) Calculate the y-component of r using the Pythagorean theorem:
Fy = Fsin 30°
Substituting for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
