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Zina [86]
4 years ago
11

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

s 13.2 m, with what minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top? Assume the rider is not strapped to the car.
Physics
1 answer:
Alla [95]4 years ago
6 0

To solve this problem it is necessary to apply the concepts related to gravitational force and centripetal force,

By definition the gravitational force on earth is defined under Newton's second law where,

F_g = mg

Where,

m = mass

g = Gravity acceleration

At the same time we know that the Centripetal force is equivalent to

F_c = \frac{mv^2}{r}

Where,

m = mass

v = Velocity

r = Radius

Since there is a balance between the two, you have to

F_g = F_c

mg = \frac{mv^2}{r}

Re-arrange to find the velocity we have,

v = \sqrt{rg}

v = \sqrt{13.2*9.8}

v = 11.37m/s

Therefore the minimum speed must be 11.37m/s

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Indiana jones (83.5 kg) is running 3.75 m/s when he jumps in a stationary 312 kg mine cart. what is their joint velocity afterwa
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Answer:

.7917 m/s

Explanation:

This is a conservation of momentum question. You have an object initially at rest (cart) so that object is initially at 0 momentum. Indiana Jones is 83.5 kg and running 3.75 m/s so he starts with a momentum of 313.125 kg * m/s because momentum is equal to mass * velocity. Once the person jumps in the cart, the cart and the person can be considered one object and by conservation of momentum, the momentum of the Indiana-cart system is equal to 313.125 kg * m/s. By that, we can set that momentum equal to the combined mass * joint velocity. So 313.125 = (83.5kg + 312kg) * joint velocity. Then just solve for the velocity. The answer should be smaller than the intial velocity of the person of 3.75 m/s because the mine cart is HUGE at 312kg.

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Why do clouds tend to form over land with a sea breeze and over water with a land breeze?
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an object of mass m is traveling at constant speed v in a circular path of radius r. how much work is done by the centripetal fo
vlada-n [284]

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

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the centripetal force is given by

F= mv²/r       (equation1)

Work done is given by

W = Fd          (equation 2)

d = 2π

work is done by the centripetal force on mass m during an angular displacement of 2π revolutions is given by:

to calculate work done using equation 1 in 2  we get

W = mv² d/r

 W = mv² × 2π /r J

The work done is by the centripetal force on mass m during an angular displacement of 2π revolutions mv²2π /r J

To know more about centripetal force :

brainly.com/question/13031430

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2 years ago
Represent 0.783 kg with Sl units having an appropriate prefix Express your answer to three significant figures and include the a
zysi [14]

Answer:

783 grams

Explanation:

Here mass is given in kg

Some of the prefixes of the SI units are

1 gram = 10⁻³ kilogram

1 milligram = 10⁻⁶ kilogram

1 microgram = 10⁻⁹ kilogram

1 nanogram = 10⁻¹² kilogram

The number is 0.783

Here, the only solution where the number of significant figures is three is gram. If any other prefix is chosen then the significant figures will increase

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So, 0.783 kg = 783 grams

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3 years ago
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