A ferris wheel of radius 100 feet is rotating at a constant angular speed Ï rad/sec counterclockwise. using a stopwatch, the rid
er finds it takes 5 seconds to go from the lowest point on the ride to a point q, which is level with the top of a 44 ft pole. assume the lowest point of the ride is 3 feet above ground level.
1 answer:
Refer to the figure shown below.
From the geometry,
y = 100 - (44 - 3) = 59 ft
From the Pythagorean theorem,
x² = 100² - 59² = 6519
x = 8007403 ft
Calculate the central angle, θ.
cos θ = 59/100 = 0.59
θ = 53.84° = 0.9397 radians
Calculate the arc length pq.
S = pq = 0.9394*100 = 93.94 ft
Calculate the angular velocity.
ω = (0.9397 radians)/(5 s) = 0.188 rad/s
Calculate the tangential velocity.
v = (100 ft)*(0.188 rad/s) = 18.8 ft/s
Calculate the time for 1 revolution.
T = (2π rad)/(0.188 rad/s) = 33.4 s
Answers:
The angular speed is 0.188 rad/s
The tangential speed is 18.8 ft/s
The time for one revolution is 33.4 s
From the geometry,
y = 100 - (44 - 3) = 59 ft
From the Pythagorean theorem,
x² = 100² - 59² = 6519
x = 8007403 ft
Calculate the central angle, θ.
cos θ = 59/100 = 0.59
θ = 53.84° = 0.9397 radians
Calculate the arc length pq.
S = pq = 0.9394*100 = 93.94 ft
Calculate the angular velocity.
ω = (0.9397 radians)/(5 s) = 0.188 rad/s
Calculate the tangential velocity.
v = (100 ft)*(0.188 rad/s) = 18.8 ft/s
Calculate the time for 1 revolution.
T = (2π rad)/(0.188 rad/s) = 33.4 s
Answers:
The angular speed is 0.188 rad/s
The tangential speed is 18.8 ft/s
The time for one revolution is 33.4 s
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Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
<em></em>
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
<em></em>
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
<em></em>
c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>
Answer:
82.76m
Explanation:
In order to find the distance of the steel ring to the ground, when its temperature has raised by 1°C, you first calculate the radius of the steel tube before its temperature increases.
You use the formula for the circumference of the steel ring:
(1)
C: circumference of the ring = 40000 km = 4*10^7m (you assume the circumference is the length of the steel tube)
you solve for r in the equation (1):
Next, you use the following formula to calculate the change in the length of the tube, when its temperature increases by 1°C:
(2)
L: final length of the tube = ?
Lo: initial length of the tube = 4*10^7m
ΔT = change in the temperature of the steel tube = 1°C
α: thermal coefficient expansion of steel = 13*10^-6 /°C
You replace the values of the parameters in the equation (2):
With the new length of the tube, you can calculate the radius of a ring formed with the tube. You again solve the equation (1) for r:
Finally, you compare both r and r' radius:
r' - r = 6,366,280.484m - 6,366,197.724m = 82.76m
Hence, the distance to the ring from the ground is 82.76m