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disa [49]
3 years ago
7

A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"

with an apparent weight only 50 % of their true weight.

Physics
2 answers:
Mashutka [201]3 years ago
8 0

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g 


Then 0.5* 9.8m/s² * 18m = 88.2 would be v² 

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

ddd [48]3 years ago
7 0

The speed of roller coaster car is about 9.4 m/s

\texttt{ }

<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

\large {\boxed {a = \frac{ v^2 } { R } }

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

\texttt{ }

Centripetal Force can be formulated as follows:

\large {\boxed {F = m \frac{ v^2 } { R } }

<em>F = Centripetal Force ( m/s² )</em>

<em>m = mass of Particle ( kg )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

Let us now tackle the problem !

\texttt{ }

<u><em>Complete Question:</em></u>

<em>A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light," with an apparent weight only 50% of their true weight. How fast is the coaster moving?</em>

<u>Given:</u>

radius of circular motion = R = 18 m

weight of passengers = w

apparent weight of passengers = N = 50%w

<u>Asked:</u>

speed of roller coaster car = v = ?

<u>Solution:</u>

\Sigma F = m \frac{v^2}{R}

w - N = m \frac{v^2}{R}

w - 50\% w = m \frac{v^2}{R}

50\% w = m \frac{v^2}{R}

50\% mg = m \frac{v^2}{R}

\frac{1}{2} g = \frac{v^2}{R}

v^2 = \frac{1}{2} gR

v = \sqrt { \frac{1}{2} g R }

v = \sqrt { \frac{1}{2} \times 9.8 \times 18}

v = \frac{21}{5} \sqrt{5} \texttt{ m/s}

\boxed{v \approx 9.4 \texttt{ m/s}}

\texttt{ }

<h3>Learn more</h3>
  • Impacts of Gravity : brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
  • The Acceleration Due To Gravity : brainly.com/question/4189441

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘C? The heat of vaporization of water at
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Answer:

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

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Mass of water to be evaporated

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volume of water

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= 100 / 1

100 cc

1 / 10 litres.

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