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3 years ago

Which point on the roller coaster's path represents the maximum potential energy?

Physics

2 answers:

mariarad [96]3 years ago

8 0

Answer:

Point A (Maximum height)

Explanation:

Potential energy is given as the product of mass, height and acceleration due to gravity hence expressed as mgh where m is the mass of roller coaster, h is the height and g is acceleration due to gravity. Since the mass and g are constant for a roller coaster, the main factor that will influence the magnitude of energy is the height. The higher the height, the higher the potential energy and vice versa. Using the attached image, the highest point is point A

sergeinik [125]3 years ago

8 0

Answer:a

Explanation: Potential energy is defined as mechanical energy, stored energy, or energy caused by its position. The energy that a ball has when perched at a top of a steep hill while it is about to roll down is an example of potential energy which is saying

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4 years ago

Read 2 more answers

The traffic on the freeway is moving at a constant speed of 24 m/sm/s. What distance does the traffic travel while the car is mo

ExtremeBDS [4]

Incomplete question as there is so much information is missing.The complete question is here

A car sits on an entrance ramp to a freeway, waiting for a break in the traffic. Then the driver accelerates with constant acceleration along the ramp and onto the freeway. The car starts from rest, moves in a straight line, and has a speed of 24 m/s (54 mi/h) when it reaches the end of the 120-m-long ramp. The traffic on the freeway is moving at a constant speed of 24 m/s. What distance does the traffic travel while the car is moving the length of the ramp?

Answer:

Distance traveled=240 m

Explanation:

Given data

Initial velocity of car v₀=0 m/s

Final velocity of car vf=24 m/s

Distance traveled by car S=120 m

To find

Distance does the traffic travel

Solution

To find the distance first we need to find time, for time first we need acceleration

$(V_{f})^{2}=(V_{o})^{2}+2aS\\ So\\a=\frac{(V_{f})^{2}-(V_{o})^{2} }{2S}\\ a=\frac{(24m/s)^{2}-(0m/s)^{2} }{2(120)}\\a=2.4 m/s^{2}$

As we find acceleration.Now we need to find time

$V_{f}=V_{i}+at\\t=\frac{V_{f}-V_{i}}{a}\\t=\frac{(24m/s)-(0m/s)}{(2.4m/s^{2} )}\\t=10s$

Now for distance

$Distance=velocity*time\\Distance=(24m/s)*(10s)\\Distance=240m$

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3 years ago