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3 years ago

At which position is the roller coaster car's amount of potential energy about equal to the amount of kinetic energy?

Physics

2 answers:

meriva3 years ago

7 0

It will be potential because the roller coaster is going down so it’s not giving off energy to the roller coaster.

Evgesh-ka [11]3 years ago

6 0

Answer:

The mass of the cars and people and the height of the track where the cars are located determine the potential energy in the system. Potential energy depends on an object’s mass and height and the gravitational pull on the object. A change in gravity would result in a change in potential energy. For example, if the track were located on the Moon instead of Earth, the potential energy would be different.

Explanation:

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3 years ago

The absolute pressure in water at a depth of 5m is read to be 145 kPa. Determine (a) the local atmospheric pressure, and (b) the

irga5000 [103]

Answer:

a) 95950 pascals

b) 137642.5 pascals

Explanation:

The absolute pressure (Pabs) on a fluid is:

$P_{abs}=P_{gauge}+P_{atm}$ (1)

With Pgauge the pressure due depth on the fluid and Patm the atmospheric pressure. Pgauge is equal to:

$P_{gauge}=\rho gh$ (2)

with ρ the fluid density, g the gravitational acceleration and h the depth on the fluid. Using (2) on (1) and solving for Patm:

$P_{atm}=P_{abs}-P_{gauge}=P_{abs}-\rho_{water} gh$

$P_{atm}=(145000Pa)-(1000\frac{kg}{m^{3}})(9.81\frac{m}{s^{2}})(5m)$

$P_{atm}=95950Pa$

b) Here we're going to use again (1) but now we have another value of density because it's other liquid, to know that value we should use the fact that specific gravity (S.G) for liquids is the ratio between fluid density and water density:

$S.G=\frac{\rho_{fluid}}{\rho_{water}}$

$\rho_{liquid}=S.G*\rho_{water}$

$\rho_{liquid}=(0.85)*(1000\frac{kg}{m^{3}})=850\frac{kg}{m^{3}}$

so:

$P_{abs}=\rho_{liquid} gh+P_{atm}=(850\frac{kg}{m^{3}})(9.81\frac{m}{s})(5m)+95950Pa$

$P_{abs}=137642.5 Pa$

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3 years ago

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3 years ago

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Explanation:

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3 years ago

A 5 kg ball rolling at 1.0 m/s hits a 15 kg ball at rest. The balls stick together after the collision What is the

Reika [66]

Answer:

0.25m/s

Explanation:

Given parameters

m₁ = 5kg

v₁ = 1.0m/s

m₂ = 15kg

v₂ = 0m/s

Unknown:

velocity after collision = ?

Solution:

Momentum before collision and after collision will be the same. For inelastic collision;

m₁v₁ + m₂v₂ = v(m₁ + m₂)

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5 x 1 + 15 x 0 = v (5 + 15 )

5 = 20v

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3 years ago

At which position is the roller coaster car's amount of potential energy about equal to the amount of kinetic energy?​

At which position is the roller coaster car's amount of potential energy about equal to the amount of kinetic energy?