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
Answer:
Explanation:
The magnitude of the net force exerted on q is known, we have the values and positions for and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted by on q. Then we can know the magnitude of the force exerted by about q, finally this will allow us to know the magnitude of
exerts a force on q in +y direction, and exerts a force on q in -y direction.
The net force on q is:
Rewriting for :
Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y = t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m