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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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Explanation:

Given

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3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate &lt;br /&
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Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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3 years ago
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The answer is c 1386j

This calculator is very helpful I use it on my homework

https://www.omnicalculator.com/physics/specific-heat
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For the electromagnetic wave in this problem, we have

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So if we solve the formula for B, we find the amplitude of the magnetic field:

B=\frac{E}{c}=\frac{10 V/m}{3\cdot 10^8 m/s}=3.33\cdot 10^{-8} T

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