Displacement stabilizes over time. It is known that exponentials raised to infinity approach zero, hence the system model will yield as time approaches infinity, resulting in 4x'' + e−0.1tx = 0. As time approaches infinity, we deduce that 4x'' equals zero. Consequently, upon integrating, we derive 4x' = c, and further integration leads to the conclusion 4x = cx + d.
Given: Speed of the sports car, v = 85 mph = 37.99 m/s. Radius of curvature, r = 525 m. Let normal weight be denoted as n and apparent weight as a. The apparent weight can be described by:... or... Consequently, this provides the necessary solution.
Answer:
A rock weighing 50kg should be positioned at a distance of 0.5m from the pivot of the seesaw.
Explanation:
τchild=τrock
We will utilize the formula for torque:
(F)child(d)child)=(F)rock(d)rock)
The gravitational force acts equally on both objects.
(m)childg(d)child)=(m)rockg(d)rock)
We can eliminate gravity from both sides of the equation for simplification.
(m)child(d)child)=(m)rock(d)rock)
Now employing the given masses for the rock and child. The seesaw's total length is 2 meters, with the child sitting at one end, placing them 1 meter from the center of the seesaw.
(25kg)(1m)=(50kg)drock
Solve for the distance where the rock should be positioned in relation to the seesaw's center.
drock=25kg⋅m50kg
drock=0.5m
The formula for range is:

Given values are:

where θ equals 14.1 degrees

Using the equation above,

The calculated range is 66.7 meters.
Therefore, the range is approximately 66.1 meters.
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)
