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valina [46]
3 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]3 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 1 m length of string is wrapped around a solid disk, of mass .25 kg and a radius of .30m, mounted on a frictionless axle. the
Nezavi [6.7K]

Answer:

60 rad/s

Explanation:

∑τ = Iα

Fr = Iα

For a solid disc, I = ½ mr².

Fr = ½ mr² α

α = 2F / (mr)

α = 2 (20 N) / (0.25 kg × 0.30 m)

α = 533.33 rad/s²

The arc length is 1 m, so the angle is:

s = rθ

1 m = 0.30 m θ

θ = 3.33 rad

Use constant acceleration equation to find ω.

ω² = ω₀² + 2αΔθ

ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)

ω = 59.6 rad/s

Rounding to one significant figure, the angular velocity is 60 rad/s.

8 0
3 years ago
How tightly does mass need to be compacted in order to become a black hole??? (2 words)
olga55 [171]

is it infinite density?

8 0
3 years ago
Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular
RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, \omega _i = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
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