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Luda [366]
3 years ago
15

While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

e motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 15 m, and you use your watch to find that each loop around takes 23 s.A. What is your speed?
B. What is the magnitude of your acceleration?
C. What is the ratio of your weight at the top of the ride to yourweight while standing on the ground?
D. What is the ratio of your weight at the bottom of the ride toyour weight while standing on the ground?
Physics
1 answer:
Dovator [93]3 years ago
6 0

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = \dfrac{2 \pi r}{T}

v = \dfrac{2\times \pi \times 15}{23}

v = 4.10 m/s

b) The magnitude of the acceleration is:

a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma  

apparent weight =mg - ma  

\dfrac{apparent \ weight}{true \  weight} = \dfrac{(mg - ma)}{(mg)}

=1- \dfrac{1.12}{9.8}

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}

\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}

= 1 + \dfrac{1.12}{9.8} \\ \\

= 1.114 m/s²

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P = W / t
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4 years ago
1. A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly hori
Dovator [93]

Answer:

The net acceleration of the SUV is 0.429 meters per square second due west.

Explanation:

Statement is incomplete. Description is presented below:

<em>A SUV along with 5 passengers has a mass of 3500 kg. It has a driving force of 2500 N directed along west on a perfectly horizontal road. The surface of the road exerts a resistance force of 500 N due east. At the same time a high wind is blowing a force of 500 N due east in the opposite direction of the car's drive force. Does the car has any acceleration? If yes, then what are the magnitude and direction of the car's acceleration?</em>

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\Sigma F = F - R -f = F_{net} (1)

Where:

R - Resistance force, measured in newtons.

f - Wind force, measured in newtons.

F_{net} - SUV net force, measured in newtons.

If we know that F = 2500\,N, R = 500\,N and f = 500\,N, then net force experimented by the SUV is:

F_{net} = 2500\,N-500\,N-500\,N

F_{net} = 1500\,N

The car has acceleration.

By definition of force for systems with constant mass, we calculate the acceleration of the vehicle below:

a_{net} = \frac{F_{net}}{m} (2)

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

If we know that F_{net} = 1500\,N and m = 3500\,kg, then the net acceleration of the car is:

a_{net} = \frac{1500\,N}{3500\,kg}

a_{net} = 0.429\,\frac{m}{s^{2}}

The net acceleration of the SUV is 0.429 meters per square second due west.

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Answer:

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