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2 years ago

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The amusement park ride shown above takes riders straight up a tall tower and then releases an apparatus holding seats. This app

aratus free-falls back to Earth and is stopped safely right above the ground. Which of the following indicates the magnitude of the gravitational force exerted on a rider of mass m on the way up and on the way down?

1 answer:

Natasha2012 [34]2 years ago

7 0

The gravitational force on the rider is:

- Way up: equal to mg

- Way down: equal to mg

Explanation:

The choices are missing: find them in the attached figure.

The force of gravity acting on a body is a force directed downward (towards the Earth's centre) and whose magnitude is

$F=mg$

where

m is the mass of the body

g is the acceleration due to gravity

The value of g is approximately constant near the Earth's surface and it is

$g=9.8 m/s^2$

During the ride, the mass of the rider, m, remains constant. This means that the magnitude of the gravitational force, mg, exerted on the rider remains constant during the rider.

Therefore, the correct answer is

- Way up: equal to mg

- Way down: equal to mg

Learn more about force of gravity:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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5.16 An insulated container, filled with 10 kg of liquid water at 20 C, is fitted with a stirrer. The stirrer is made to turn by

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Answer:

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b) $\Delta E=2.425kJ$

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d) Q=-2.425kJ

Explanation:

a)

First of all, we need to do a drawing of what the system looks like, this will help us visualize the problem better and take the best possible approach. (see attached picture)

The problem states that this will be an ideal system. This is, there will be no friction loss and all the work done by the object is transferred to the water. Therefore, we need to calculate the work done by the object when falling those 10m. Work done is calculated by using the following formula:

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Where:

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F= force applied [N]

d= distance [m]

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F=mg

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d=h

so the formula gets the following shape:

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so now e can substitute:

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which yields:

W=2.425kJ

b) Since all the work is tansferred to the water, then the increase in internal energy will be the same as the work done by the object, so:

$\Delta E=2.425kJ$

c) In order to find the final temperature of the water after all the energy has been transferred we can make use of the following formula:

$\Delta Q=mC_{p}(T_{f}-T_{0})$

Where:

Q= heat transferred

m=mass

$C_{p}$ =specific heat

$T_{f}$ = Final temperature.

$T_{0}$ = initial temperature.

So we can solve the forula for the final temperature so we get:

$T_{f}=\frac{\Delta Q}{mC_{p}}+T_{0}$

So now we can substitute the data we know:

$T_{f}=\frac{2 425J}{(10000g)(4.1813\frac{J}{g-C})}+20^{o}C$

Which yields:

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d)

For part d, we know that the amount of heat to be removed for the water to reach its original temperature is the same amount of energy you inputed with the difference that since the energy is being removed this means that it will be negative.

$\Delta Q=-2.425kJ$

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