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sveticcg [70]
3 years ago
7

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.

Physics
1 answer:
Zinaida [17]3 years ago
4 0
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

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Specific Gravity of the fluid = 1.25 
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 Density of the Fluid p = 78 lbm/ft^3
 Difference in pressure as we got the differential height, dP = p x g x h  dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
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 (a) Pressure in the arm that is at Higher 
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3 years ago
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alina1380 [7]
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grandymaker [24]
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3 years ago
The atlas stone’s is a strong man competition where athletes have to load 5 stones of masses 100kg, 120kg, 140kg, 160kg and 180k
lyudmila [28]

Answer: 6250 joules

Explanation:

The work needed to lift an object of mass M by a height H is equal to:

w = M*g*H

where h = 10m/s^2

then the total work that he did is equal to the sum of the work for every stone:

W = (100kg*g*H) + (120kg*g*H) + (140kg*g*H) + (160kg*g*H) + (180kg*g*H)

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= (500kg)*g*H

and now we can repalce g by 10m/s^2 and H by 125cm

But you can notice that we have two different units of distance, so knowing that 100cm = 1m

we can write H =  125cm = (125/100) m = 1.25 m

Then we have:

H = 500kg*10m/s^2*1.25m = 6250 J

3 0
3 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
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