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3 years ago

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A student has a weight of 655 N. While riding a roller coaster they seem to weigh 1.96x 103 N at the bottom of a dip that has a

radius of 18.0 m. What is the speed of the roller coaster at this point?

2 answers:

kherson [118]3 years ago

8 0

Answer:

Explanation:

weight, mg = 655 N

m = 655 / 9.8 = 66.84 Kg

N = 1.96 x 10^3 N

radius, r = 18 m

Let v be the speed of roller coaster.

So, the apparent weight

N = mg + mv²/r

1.96 x 1000 = 655 + 66.84 v² / 18

1305 = 3.71 v²

v² = 351.75

v = 18.76 m/s

Nadya [2.5K]3 years ago

4 0

Answer:

The speed of the roller coaster at this point is 18.74 m/s.

Explanation:

Given that,

Weight of the student, W = 655 kg

Weight of the roller coaster, $F=1.96\times 10^3\ N$

Radius of the roller coaster, r = 18 m

At the bottom of the loop, the weight of the roller coaster us given by :

$F=W+\dfrac{mv^2}{r}$

If m is the mass of the roller coaster,

$W=mg$

$m=\dfrac{W}{g}$

$m=\dfrac{655}{9.8}$

m = 66.83 kg

So,

$F=W+\dfrac{mv^2}{r}$

$v=\sqrt{\dfrac{(F-W)r}{m}}$

$v=\sqrt{\dfrac{(1.96\times 10^3-655)\times 18}{66.83}}$

v = 18.74 m/s

So, the speed of the roller coaster at this point is 18.74 m/s. Hence, this is the required solution.

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a) $s_{T} = 30\,m$ , b) $t = 5\,min$ , c) $\Delta t = 6.667\,s$ , d) $\Delta s_{R} = 33.333\,m$ , e) $t' = 11.667\,s$ , f) The rabbit won the race.

Explanation:

a) As turtle moves at constant speed, its position is determined by the following formula:

$s_{T} = v_{T}\cdot t$

Where:

$t$ - Time, measured in seconds.

$v_{T}$ - Velocity of the turtle, measured in meters per second.

$s_{T}$ - Position of the turtle, measured in meters.

Then, the position of the turtle when the rabbit starts to run is:

$s_{T} = \left(0.5\,\frac{m}{s} \right)\cdot (60\,s)$

$s_{T} = 30\,m$

The position of the turtle when the rabbit starts to run is 30 meters.

b) The time needed for the turtle to finish the race is:

$t = \frac{s_{T}}{v_{T}}$

$t = \frac{150\,m}{0.5\,\frac{m}{s} }$

$t = 300\,s$

$t = 5\,min$

The time needed for the turtle to finish the race is 5 minutes.

c) As rabbit experiments a constant acceleration until maximum velocity is reached and moves at constant speed afterwards, the time required to reach such speed is:

$v_{R} = v_{o,R} + a_{R}\cdot \Delta t$

Where:

$v_{R}$ - Final velocity of the rabbit, measured in meters per second.

$v_{o,R}$ - Initial velocity of the rabbit, measured in meters per second.

$a_{R}$ - Acceleration of the rabbit, measured in $\frac{m}{s^{2}}$ .

$\Delta t$ - Running time, measured in second.

$\Delta t = \frac{v_{R}-v_{o,R}}{a_{R}}$

$\Delta t = \frac{10\,\frac{m}{s}-0\,\frac{m}{s}}{1.50\,\frac{m}{s^{2}} }$

$\Delta t = 6.667\,s$

The time taken by the rabbit to reach maximum speed is 6.667 s.

d) On the other hand, the position reached by the rabbit when maximum speed is reached is determined by the following equation of motion:

$v_{R}^{2} = v_{o,R}^{2} + 2\cdot a_{R}\cdot \Delta s_{R}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

$\Delta s_{R} = \frac{v_{R}^{2}-v_{o,R}^{2}}{2\cdot a_{R}}$

Where $\Delta s_{R}$ is the travelled distance of the rabbit from rest to maximum speed.

$\Delta s_{R} = \frac{\left(10\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(1.50\,\frac{m}{s^{2}} \right)}$

$\Delta s_{R} = 33.333\,m$

The distance travelled by the rabbit from rest to maximum speed is 33.333 meters.

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$t' = \frac{\Delta s_{R}}{v_{R}}$

$t' = \frac{150\,m-33.333\,m}{10\,\frac{m}{s} }$

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$t_{T} = 300\,s$

Rabbit:

$t_{R} = 60\,s + 6.667\,s + 11.667\,s$

$t_{R} = 78.334\,s$

The rabbit won the race as $t_{R} < t_{T}$ .

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