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Vesnalui [34]
2 years ago
14

Students are asked to create roller coasters for marbles. Their goal is to design a coaster with the tallest possible hill that

a marble released from a height of 1.5 m (meters) can clear. The marbles will experience some air resistance and friction as
they move.

What should the students keep in mind as they build their designs?

a)The hill can be taller than 1.5 m (meters), because the marble will be moving faster than its initial velocity allowing it to travel higher than its release height.

b)The hill can be taller than 1.5 m (meters), because the marble will gain mechanical energy as it moves allowing it to travel higher than its release height.

c)The hill should be a little less than 1.5 m (meters) high, because the marble will lose mechanical energy as it moves, preventing it from reaching its release
height.

d)The hill can be exactly 1.5 m (meters) high, because mechanical energy is always
conserved allowing the marble to travel to its release height.
Physics
1 answer:
ivann1987 [24]2 years ago
3 0

Answer:

Kinetic Energy.

Explanation:

The movement of a roller coaster is accomplished by the conversion of potential energy to kinetic energy. The roller coaster cars gain potential energy as they are pulled to the top of the first hill. As the cars descend the potential energy is converted to kinetic energy.

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A 1.20-m cylindrical rod of diameter 0.570 cm is connected to a power supply that maintains a constant potential difference of 1
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(a) 1.72\cdot 10^{-5} \Omega m

The resistance of the rod is given by:

R=\rho \frac{L}{A} (1)

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A=\pi r^2=\pi (2.85\cdot 10^{-3} m)^2=2.55\cdot 10^{-5} m^2

The resistance at 20°C can be found by using Ohm's law. In fact, we know:

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(b) 8.57\cdot 10^{-4} /{\circ}C

First of all, let's find the new resistance of the wire at 92.0°C. In this case, the current is

I = 17.5 A

So the resistance is

R=\frac{V}{I}=\frac{15.0 V}{17.5 A}=0.86 \Omega

The equation that gives the change in resistance as a function of the temperature is

R(T)=R_0 (1+\alpha(T-T_0))

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R(T)=0.86 \Omega is the resistance at the new temperature (92.0°C)

R_0=0.81 \Omega is the resistance at the original temperature (20.0°C)

\alpha is the temperature coefficient of resistivity

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T_0 = 20^{\circ}

Solving the formula for \alpha, we find

\alpha=\frac{\frac{R(T)}{R_0}-1}{T-T_0}=\frac{\frac{0.86 \Omega}{0.81 \Omega}-1}{92C-20C}=8.57\cdot 10^{-4} /{\circ}C

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