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riadik2000 [5.3K]
2 years ago
12

An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an

d continues up the next hill. The end of the roller coaster has a level surface that is 4m off the ground. Assume there is no friction on the roller coaster ridea and energy is concerved.
Q. How tall is the hill if the roller coaster is traveling at 10m/s at the top of it?
Q. What will be the velocity of the roller coaster at the top of the loop?
Physics
1 answer:
aalyn [17]2 years ago
6 0
Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H)  =  (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) =<em> </em><u><em>359,200 joules</em></u><em> </em>of energy.

There's no friction, so it'll have <u>that same energy</u> at every point of the story.
=================================

Skip the loop for a moment, because the first question concerns the hill after
the loop.  We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = <em>40.71 meters</em>
=================================

Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = <em>21.61 m/s</em>
===============================

Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground,  PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic.  KE =  (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = <em>28.63 m/s</em> at the end
=======================================

If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere.  Sorry.
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Answer:

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Explanation:

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Describe how one plays Dr.Dogeball​
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2 years ago
Which planet in the list below is larger than Earth? A. Saturn B. Mars C. Mercury D. Venus
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Read 2 more answers
A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
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Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

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Dua titik yang bermuatan listrik yang sama, mula-mula berjarak 5 cm dan saling tarik menarik dengan gaya 225 N. Agar kedua titik
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Answer:

15 cm

Explanation:

Dari pertanyaan yang diberikan di atas, diperoleh data sebagai berikut:

Gaya 1 (F₁) = 225 N

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Gaya 2 (F₂) = 25 N

Jarak terpisah 2 (d₂) =?

Kita dapat memperoleh persamaan yang berkaitan dengan gaya dan jarak muatan dua titik dengan menggunakan rumus berikut:

F = Kq₁q₂ / d²

Perbanyak silang

Fd² = Kq₁q₂

Menjaga Kq₁q₂ konstan, kita memiliki:

F₁d₁² = F₂d₂²

Dengan rumus di atas maka diperoleh jarak sebagai berikut:

Gaya 1 (F₁) = 225 N

Jarak terpisah 1 (d) = 5 cm

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Jarak terpisah 2 (d₂) =?

F₁d₁² = F₂d₂²

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225 × 25 = 25 × d₂²

5625 = 25 × d₂²

Bagilah kedua sisinya dengan 25

d₂² = 5625/25

d₂² = 225

Hitung akar kuadrat dari kedua sisi

d₂ = √225

d₂ = 15 cm

Oleh karena itu, muatan dua titik harus berjarak 15 cm untuk memiliki gaya tarik 25 N

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