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sveticcg [70]
2 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]2 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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Yes this is not gonna work
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2 years ago
An electric device, which heats water by immersing a resistance wire in the water, generates 50 cal of heat per second when an e
Oksi-84 [34.3K]

Answer:

0.69 ohm

Explanation:

Heat generated per second, H = 50 cal/s

Potential difference, V = 12 V

Let R is the resistance of coil.

The formula for the heat is given by

H = \frac{V^{2}}{R}t

50\times 4.186 = \frac{12^{2}}{R}\times 1

R = 0.69 ohm

3 0
2 years ago
A cart is driven by a large propeller or fan, which can accelerate or decelerate the cart. The cart starts out at the position x
mash [69]

Answer:

The acceleration of the cart is 1.0 m\s^2 in the negative direction.

Explanation:

Using the equation of motion:

Vf^2 = Vi^2 + 2*a*x

2*a*x = Vf^2 - Vi^2

a = (Vf^2 - Vi^2)/ 2*x

Where Vf is the final velocity of the cart, Vi is the initial velocity of the cart, a the acceleration of the cart and x the displacement of the cart.

Let x = Xf -Xi

Where Xf is the final position of the cart and Xi the initial position of the cart.

x = 12.5 - 0

x = 12.5

The cart comes to a stop before changing direction

Vf = 0 m/s

a = (0^2 - 5^2)/ 2*12.5

a = - 1 m/s^2

The cart is decelerating

Therefore the acceleration of the cart is 1.0 m\s^2 in the negative direction.

5 0
2 years ago
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
2 years ago
A Hooke's law bowstring is stretched x meters until a force of f newtons is applied, and then held. By what factor will the elas
Liono4ka [1.6K]

The elastic potential energy increases by a factor of 9

Explanation:

The elastic potential energy of a bowstring is given by

E=\frac{1}{2}kx^2 (1)

where

k is the spring constant

x is the elongation of the bowstring

Hooke's law states the relationship between the force applied and the elongation of an elastic object:

F=kx

where

F is the force applied

x is the elongation

We can rewrite it as

x=\frac{F}{k}

And substituting into (1),

E=\frac{1}{2}k(\frac{F}{k})^2=\frac{F^2}{2k}

In  this problem, the force applied to the bowstring is tripled,

F' = 3F

So the final elastic potential energy is:

E'=\frac{(3F)^2}{2k}=9(\frac{F^2}{2k})=9E

So, the elastic potential energy increases by a factor of 9.

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

7 0
2 years ago
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