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valina [46]
2 years ago
14

Students were asked to create roller coasters for marbles. The only requirement is that the roller coaster include at least one

hill that the marble must roll over in order for the roller coaster to be considered a success. Students are building their designs with marbles and foam tubing so there will be some air resistance and friction. What should the students keep in mind if they want to create a successful roller coaster?
Question 23 options:

1)


Without an extra push at the bottom of the first hill, there is no way the car will make it back up a hill of any height because the marble doesn't have enough mass or velocity

2)


The hill should be at least a little lower than the starting height because some of the kinetic energy at the bottom of the first hill will be converted to other types of energy due to friction so it will not have as much potential energy at the top of the next hill

3)

The hill should be taller than the starting height because the marble will pick up speed on the downward hill and the increased velocity will allow it to travel higher on the next hill

4)


The hill should be the exact same height as the starting height because 100% of the kinetic energy at the bottom of the first hill will be converted back to potential energy at the top of the next hill
Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Since energy is lost in the roller coaster due to friction, the hill should little lower than the starting height since some of the kinetic energy at the bottom of the first hill is lost  due to friction so it will not have as much potential energy at the top of the next hill.

A roller coaster is a good way to demonstrate the principle of conservation  of energy. Recall that energy is neither created nor destroyed but can be converted from one form to another.

In a roller coaster, all the heights are not the same because energy is lost along the line. Therefore, the students must bear in mind that  the hill should be at least a little lower than the starting height because some of the kinetic energy at the bottom of the first hill will be converted to other types of energy due to friction so it will not have as much potential energy at the top of the next hill.

Learn more: brainly.com/question/14281129

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A pair of narrow slits, separated by 1.8 mm, is illuminated by a monochromatic light source. Light waves arrive at the two slits
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Answer:

750 nm

Explanation:

d  = separation of the slits = 1.8 mm = 0.0018 m

λ = wavelength of monochromatic light

D  = screen distance = 4.8 m

y = position of first bright fringe = \frac{1cm}{5 fringe} = \frac{0.01}{5} = 0.002 m

n  = order = 1

Position of first bright fringe is given as

y = \frac{nD\lambda }{d}

0.002 = \frac{(1)(4.8)\lambda }{0.0018}

λ = 7.5 x 10⁻⁷ m

λ = 750 nm

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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr
mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = 5 \ * (M \ _{sun})

where : M_{sun} = 1.99*10^{33} \ kg

Then; we have:

 m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear m_{ear} = 0.030 \ kg

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R -  d) \omega^2

\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)

The net force on the ear farther to the black hole is :

\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R +  d) \omega^2

\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

T = \frac{3GMm_{ear}d}{R^3}

T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}

T = 2073.9 N\\T = 2.07 KN

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

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Answer:

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