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3 years ago

Wind tends to move from pressure areas. low, high high, high low, low high, low

Physics

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

high, low

Explanation:

Energy always flows from a higher level to a lower level.
It is analogous to the waterfall where waterfalls from a higher level to a lower level.
So in the case of the pressure of the gas, when there are any numbers of molecules in a given volume of space. The gas is said to be at high pressure.
When there are fewer molecules in the given volume. The gas is said to be at lower pressure.
Due to a large number of atoms, the high-pressure gas exerts more force on the container than the force exerted by the low-pressure gas.
If a hose is connected between these two containers, gas rushes from high pressure to the low pressure. Since the force exerted by the high-pressure gas is greater than that of low-pressure gas.

So, the wind tends to move from high-pressure areas to low pressure.

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What is the change in internal energy if 60 J of heat is released from a system and 30 J of work is done on the system? Use U =

k0ka [10]

The change in internal energy of the system is +30 J

Explanation:

We can solve this problem by using the first law of thermodynamics, which states that the change in internal energy of a system is given by the equation:

$\Delta U = Q -W$

where

$\Delta U$ is the change in internal energy

Q is the heat absorbed by the system (positive if it is absorbed, negative if it is released)

W is the work done by the system (positive if it is done by the system, negative if it is done by the surroundings on the system)

Therefore, in this problem, we have

$Q=-60 J$ (heat released by the system)

$W=-30 J$ (work done on the system)

Therefore, the change in internal energy is

$\Delta U = -60 - (-30) = +30 J$

Learn more about thermodynamics:

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3 years ago

Problem:

pav-90 [236]

Answer:

Total energy is constant

Explanation:

The laws of thermodynamics state that thermal energy (heat) is always transferred from a hot body (higher temperature) to a cold body (lower temperature).

This is because in a hot body, the molecules on average have more kinetic energy (they move faster), so by colliding with the molecules of the cold body, they transfer part of their energy to them. So, the temperature of the hot body decreases, while the temperature of the cold body increases.

This process ends when the two bodies reach the same temperature: we talk about thermal equilibrium.

In this problem therefore, this means that the thermal energy is transferred from the hot water to the cold water.

However, the law of conservation of energy states that the total energy of an isolated system is constant: therefore here, if we consider the hot water + cold water as an isolated system (no exchange of energy with the surroundings), this means that their total energy remains constant.

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3 years ago

A wire carries a steady current of 2.20

alina1380 [7]

By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.

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3 years ago

A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout

nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
So, we can write the following equation:

$E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)$

As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
So, on the outer surface of the shell there must be a charge that be the difference between them:

$Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C$

Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

$E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C$

As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.

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3 years ago

Which best describes the current atomic theory?

ivolga24 [154]

Choice-C is a correct statement.

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3 years ago

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