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Sauron [17]
3 years ago
11

The bar graph shows energy data taken from a roller coaster at a theme park. analyze the data and assess its validity. 3-5 sente

nces, record your conclusions.

Physics
1 answer:
hichkok12 [17]3 years ago
4 0

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The data shown by the bar graph is "valid".

According to the law of conservation of energy, the total energy of the system must remain constant at any given point. Hence, the sum of the kinetic energy and the potential energy of the roller coaster must be constant at any given time.

Considering the bottom of the first hill:

Total Energy at 1st Hill = Kinetic Energy + Potential Energy

From the data given in the bar graph:

Total Energy at 1st Hill = 2,500,000 J + 0 J (since at the bottom potential energy is zero due to zero height)

Total Energy at 1st Hill = 2,500,000 J

Now, considering the top of the second hill:

Total Energy at 2nd Hill = Kinetic Energy + Potential Energy

From the data given in the bar graph:

Total Energy at 2nd Hill = 1,000,000 J + 1,500,000 J

Total Energy at 2nd Hill = 2,500,000 J

Hence,

Total Energy at 1st Hill = Total Energy at 2nd Hill

Therefore, the given bar graph is "valid".

Learn more about the law of conservation of energy here:

brainly.com/question/20971995?referrer=searchResults

The attached picture explains the law of conservation of energy.

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Use these images to identify each state of matter. O O CC 0 COM A B C​
Nikolay [14]

Answer:

A=Gas

B= Solid

C=Liquid

Explanation:

A gas has the particles dispersed everywhere

so A is the gas

Liquid and Solid have them more close together

but the solid has them in some order

so as you can see

B is solid

and

C is liquid

7 0
3 years ago
What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the pro
Novay_Z [31]

This question is incomplete, the complete question is;

A weightlifter holds a 1,300 N barbell 1 meter above the ground. One end of a 2-meter-long chain hangs from the center of the barbell. The chain has a total weight of 400 N. How much work (in J) is required to lift the barbell to a height of 2 m?

What is the average force (average with respect to height of the barbell from the ground) exerted by the weightlifter in the process?

Answer: Average force exerted by the weightlifter in the process = 1600N

Explanation:

To find Work done to lift a barbell and half of the hanging chain we say;

W₁ = ( 1300N + (1/2 × 400N)) × 1m

W₁ = (1300 + 200) Nm

W₁ = 1500J

now work done to lift the upper half of the chain we say:

W₂ = (1/2 × 400N) ×  (1/2 × 1m)

W₂ = 200N × 0.5m

W₂ = 100J

So total work done will be

W = W₁ + W₂

W = 1500J + 100J

W = 1600J

To find the average force exerted by the weight lifter, we say;

F = W/D

F = (1600 / 1m) N

F = 1600N

∴Average force = 1600N

6 0
3 years ago
If the magnitude of b⃗ increases while its direction remains unchanged, how will the magnetic flux through the coil change?
Lapatulllka [165]
Flux = B * Area
If B increases, flux also increases.
4 0
3 years ago
A girl is pedaling her bicycle at a speed of 0.10 km/min. What will her displacement be after two
Naily [24]

Answer:

im not so sure about the answer but in my opioion it would be 12

Explanation:

reason being is you multiply 0.10 km/minute and in 2 hours there is 120 minutes which the answer leads to 12 km/min

6 0
3 years ago
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
3 years ago
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