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Brrunno [24]
3 years ago
8

A container of gas is held at a constant volume. Which will most likely happen to the temperature if the pressure of the gas inc

reases?
The temperature will increase at the same rate as the pressure increases.

The temperature will decrease at the same rate as the pressure increases.

The temperature will increase at half the same rate as the increase in pressure.

The temperature will decrease at half the same rate as the increase in pressure.​
Physics
1 answer:
sergij07 [2.7K]3 years ago
5 0

Answer:

With more particles there will be more collisions and so a greater pressure. The number of particles is proportional to pressure, if the volume of the container and the temperature remain constant. ... This happens when the temperature is increased.

Explanation:

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Which statements describe the energy in a food web? Select three options.
hichkok12 [17]

Answer:

it loses engry  it follows difernt paths

Explanation:

8 0
3 years ago
Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f
tatiyna

Answer:

a) h=3.16 m, b)  v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

           Em₀ = U = mg h

final point. Lowest point

           Em_{f} = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

           K = ½ m v_{cm }^{2} + ½ I_{cm} w²

angular and linear speed are related

           v = w r

           w = v / r

            K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

            Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

             Em₀ = Em_{f}

             mg h = ½ v_{cm }^{2} (m + I_{cm} / r²)         (1)

             h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

             I_{cm} = ⅔ m r²

we substitute

            h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

            h = ½ v_{cm }^{2}/g    5/3

             h = 5/6 v_{cm }^{2} / g

           

let's calculate

           h = 5/6 6.1 2 / 9.8

           h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

          v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

              I_{cm} = ½ m r²

we substitute

             v_{cm } = √ [2gh / (1 + ½)]

             v_{cm } = √(4/3 gh)

let's calculate

             v_{cm } = √ (4/3 9.8 3.16)

             v_{cm }^ = 6.43 m / s

4 0
3 years ago
Light incident on a lake surface is partly reflected and partly refracted.What is the differences between the reflected ray and
iren2701 [21]

Answer: As per the question, a ray of light is incident on a surface and it is partly reflected and refracted. The incident light is an unpolarised light. The reflected light is partially polarised.

If the angle of incidence becomes equal to the Brester angle (polarising angle), then the reflected light is completely plane polarised.

5 0
3 years ago
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there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f
Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

4 0
3 years ago
Part of the plant that isresponsible for the transport of water from the roots to the leaves​
nydimaria [60]

Answer:

if you mean *responsible* for the transport of water from the roots to leaves is Xylem

8 0
3 years ago
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