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SVEN [57.7K]
2 years ago
6

Three forces that act on a roller coaster:

Physics
2 answers:
Gelneren [198K]2 years ago
7 0
There is also potential and kenetic energy
Andre45 [30]2 years ago
6 0
Yes, all of these could be applied to a roller coaster.
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How to do this question
Anna71 [15]

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop.  There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

4 0
3 years ago
A tennis ball is dropped from a height of 1.21m above the ground. Calculate its velocity when it is 0.27m from the ground.
wlad13 [49]
9.81m/s^2 x .27 = 2.6487m/s
8 0
3 years ago
Following are the different layers of the sun’s atmosphere. Rank them based on the order in which a probe would encounter them w
blsea [12.9K]

Answer:

Going from earth to the sun a probe would encounter the next layers in order:

  • Corona
  • Transition Region
  • Chromosphere
  • Photosphere
  • Convection Zone
  • Radiative Zone
  • Core

A brief description of them:

Corona is the outermost layer and it cannot  be seen with the naked eye, is starts at about 2100 km from the surface of the sun and it has no limit defined.

Transition Region is between the corona and the chromosphere, it has an extension of about 100km

The chromosphere is between 400 km from the surface of the sun to 2100 km. In this layer the further you get away from the sun it gets hotter.

The photosphere is the surface of the sun, the part that we can see, and extends from the surface to 400km.

The convection zone is where convection happens, hot gas rises, cools and rises again.

Radiative Zone is where the photons try to rise to move to higher layers.

The core of the Sun is where nuclear fusion occurs due to the very high temperatures.

6 0
3 years ago
What explains the key difference between a bomb calorimeter and a coffee cup calorimeter?.
Virty [35]

The key difference between a bomb calorimeter and a coffee cup calorimeter is high temperature.

<h3>What is bomb calorimeter?</h3>

A bomb calorimeter is an apparatus that can measure heats of combustion, used in various applications such as calculating the calorific value of foods and fuels.

<h3>What is coffee cup calorimeter?</h3>

A coffee cup calorimeter is a cup used to provide insulation when materials are mixed inside of it.

<h3>Difference between the two calorimeter</h3>
  • The coffee cup calorimeter can't be used for high-temperature reactions, either, because they would melt the cup.
  • A bomb calorimeter is used to measure heat flows for gases and ​high-temperature reactions

Learn more about calorimeter here: brainly.com/question/1407669

#SPJ1

3 0
2 years ago
A stone is tied to a string and whirled at constant speed in a horizontal circle. The speed is then doubled without changing the
Tanya [424]

Answer:

Afterward the magnitude of the acceleration of the stone is four times as great

Explanation:

A stone tied to a string and whirled at a constant speed in a horizontal circle will have a centripetal acceleration given by

a = \frac{v^{2} }{r}

Where a is the centripetal acceleration

v is the speed

and r is the radius of the circle

Here, the radius of the circle is the length of the string.

Now, from the question

The speed is then doubled without changing the length of the string,

Let the new speed be v_{2}, that is

v_{2} = 2v

and without changing the length of the string means radius r is constant.

To determine the magnitude of the acceleration of the stone afterwards,

Let the new acceleration be a_{2}.

Then we can write that

a_{2} = \frac{v_{2}^{2}  }{r}

From

a = \frac{v^{2} }{r}

v = \sqrt{ar}

Recall that

v_{2} = 2v

∴ v_{2} = 2\sqrt{ar}

Now, we will put the value of v_{2} into

a_{2} = \frac{v_{2}^{2}  }{r}

Then,

a_{2} = \frac{(2\sqrt{ar}) ^{2}  }{r}

a_{2} = \frac{4ar }{r}

a_{2} = 4a

The new magnitude of the acceleration of the stone is four times the initial acceleration.

Hence,

Afterward the magnitude of the acceleration of the stone is four times as great

3 0
3 years ago
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