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Elenna [48]
3 years ago
11

Explain why the roller coaster’s potential energy is greater at point 1 than at point 4.

Physics
2 answers:
Virty [35]3 years ago
3 0

Potential energy is directly related to the height of an object. Point 1 is higher than point 4, so the potential energy is greater at point 1.

spayn [35]3 years ago
3 0

The most probable answer would be Point 1 is higher than Point 4.


Potential gravitational energy is computed by the formula:


PEgrav =mgh

m = mass

g = acceleration due to gravity (this is a constant value of 9.8m/s²)

h = height


The roller coaster of course, no matter where in the track it is would have the same mass (Unless someone falls off -_-) Pull of gravity is the same as well.


So the only factor that would affect it is the height. So if the height goes up, the potential energy goes up too. If the height goes down, then the potential energy goes down too.


Hopefully you'll come up with the right answer with this. v(0-0)v


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Nio conducts a science experiment. He puts yeast into a bottle with water and sugar, and then caps the bottle with a balloon. He
KATRIN_1 [288]

Answer: Option (C) is the correct answer.

Explanation:

When yeast is added into a bottle with water and sugar then after some time yeast becomes activated. This results in the formation of a chemical change as bubbles are formed on the surface of the mixture in the bottle.

This indicates that a gas has been released due to the chemical change.

Thus, we can conclude that most likely a chemical change has occurred because a gas was produced.

3 0
3 years ago
Read 2 more answers
I have a problem in this questions?
photoshop1234 [79]

Answer:

8.46E+1

Explanation:

From the question given above, the following data were obtained:

Charge 1 (q₁) = 39 C

Charge 2 (q₂) = –53 C

Force (F) of attraction = 26×10⁸ N

Electrical constant K) = 9×10⁹ Nm²/C²

Distance apart (r) =?

The distance between the two charges can be obtained as follow:

F = Kq₁q₂ / r²

26×10⁸ = 9×10⁹ × 39 × 53 / r²

26×10⁸ = 1.8603×10¹³ / r²

Cross multiply

26×10⁸ × r² = 1.8603×10¹³

Divide both side by 26×10⁸

r² = 1.8603×10¹³ / 26×10⁸

r² = 7155

Take the square root of both side

r = √7155

r = 84.6 m

r = 8.46E+1 m

7 0
2 years ago
Calculate the separation between the two lowest levels for an O2 molecule in a one-dimensional container of length 5.0 cm. At wh
MariettaO [177]

Answer:

The separation between the two lowest levels =  1.24 * 10^{-39}J

The values of n where the energy of molecule reaches 1/2 kT at 300K = 2.2 * 10^{9}

The separation at this level = 1.8 * 10^{-30}J

Explanation:

Knowing the formula

En = \frac{n^{2} h^{2}  }{8 mL^{2} }

Mass of oxygen molecule

m (O2) = 32 amu * \frac{1.6605 * 10^{-27 kg} }{1 amu}

So the energy diference between the two lowest levels:

E2 - E1 = \frac{3h^{2} }{8mL^{2} }

E2 - E1 =  \frac{3 * (6.626 * 10^{-34} Js)^{2} } {8 * 32 amu * (\frac{1.6605 * 10^{-27 kg} }{1 amu})* (5*10^{-2})^{2}   } = 1.24 * 10^{-39}J

Now we should find n where the energy of molecule reaches 1/2 kT

En = \frac{n^{2} h^{2}  }{8 mL^{2} } = \frac{1}{2}kT

\frac{h^{2}  }{8 mL^{2} } = 4.13 * 10^{-14}J

n^{2} *  (4.13 * 10^{-14}J) = \frac{1}{2} (1.38 * 10^{-23}JK^{-1}) * 300K

n = 2.2 * 10^{9}

by the end is necessary to calculate the separation of the level

En - En-1 = (n^{2} - (n - 1)^{2}) * \frac{h^{2}  }{8 mL^{2} }

              = 1.8 * 10^{-30}J

4 0
3 years ago
A segment of wire carries a current of 25 A along the x axis from x = −2.0 m to x = 0 and then along the y axis from y = 0 to y
Rashid [163]

Answer: F = 2N

Explanation: If a current i is flowing in a wire of length L lying in a region of magnetic field B, then the magnetic force acting on the wire is given by

F = BIL

Please find the attached file for the solution

8 0
3 years ago
A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha
ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, L=400\ \mu H=400\times 10^{-6}\ H

Capacitance of parallel LCR circuit, C=0.3\ \mu F=0.3\times 10^{-6}\ F

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

f=\dfrac{1}{2\pi \sqrt{LC} }

f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }

f=14528.79\ Hz

or

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0
3 years ago
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