Answer:
A = 2.36m/s
B = 3.71m/s²
C = 29.61m/s2
Explanation:
First, we convert the diameter of the ride from ft to m
10ft = 3m
Speed of the rider is the
v = circumference of the circle divided by time of rotation
v = [2π(D/2)]/T
v = [2π(3/2)]/4
v = 3π/4
v = 2.36m/s
Radial acceleration can also be found as a = v²/r
Where v = speed of the rider
r = radius of the ride
a = 2.36²/1.5
a = 3.71m/s²
If the time of revolution is halved, then radial acceleration is
A = 4π²R/T²
A = (4 * π² * 3)/2²
A = 118.44/4
A = 29.61m/s²
The movements of the tectonic plates
Answer:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:
![qvB = m\frac{v^2}{r}](https://tex.z-dn.net/?f=qvB%20%3D%20m%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is
![T=\frac{2\pi r}{v}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%5Cpi%20r%7D%7Bv%7D)
Re-arranging for r
![r=\frac{Tv}{2\pi}](https://tex.z-dn.net/?f=r%3D%5Cfrac%7BTv%7D%7B2%5Cpi%7D)
And substituting into the previous equation
![qvB = m \frac{Tv^3}{2\pi}](https://tex.z-dn.net/?f=qvB%20%3D%20m%20%5Cfrac%7BTv%5E3%7D%7B2%5Cpi%7D)
Solving for T,
![T=\frac{2\pi q B}{m v^2}](https://tex.z-dn.net/?f=T%3D%5Cfrac%7B2%5Cpi%20q%20B%7D%7Bm%20v%5E2%7D)
So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
Answer:
a) 600 meters
b) between 0 and 10 seconds, and between 30 and 40 seconds.
c) the average of the magnitude of the velocity function is 15 m/s
Explanation:
a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:
![Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m](https://tex.z-dn.net/?f=Area%20%5C%2C%5C%2CTrapezoid%3D%28%5Cleft%5BB%2Bb%5D%5C%2C%28H%2F2%29%5C%5Cdisplacement%3D%20%5Cleft%5B%2840-0%29%2B%2830-10%29%5Cright%5D%20%5C%2C%2820%2F2%29%3D600%5C%2C%5C%2Cm)
b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.
Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.
c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:
![Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx](https://tex.z-dn.net/?f=Average%5C%2Cof%5C%2Cf%28x%29%3D%5Cfrac%7B1%7D%7Bb-a%7D%20%5Cint%5Climits%5Eb_a%20%7Bf%28x%29%7D%20%5C%2C%20dx)
Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:
![Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}](https://tex.z-dn.net/?f=Avearage%3D%5Cfrac%7B600%5C%2C%5C%2Cm%7D%7B40%5C%2C%5C%2Cs%7D%3D%2015%5C%2C%5C%2C%5Cfrac%7Bm%7Ds%7D)
A. Physics has changed the course of the world.