What change would occur in the kinetic energy of vapor if a container of vapor is placed at 0 Kelvin?

Physics

1 answer:

n200080 [17]3 years ago

7 0

Answer:

The kinetic energy of gas particles will increase. The kinetic energy of gas particles will decrease. The kinetic energy of gas particles will remain unchanged

Explanation:

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Which device uses a rotating magnetic field to produce an electric current?

Kay [80]

The correct answer is C) a motor.
In particular, we are talking about an AC motor, which produces an alternating current. In an AC motor, a coil is immersed in a rotating magnetic field. Due to the motion of the magnetic field,the angle between the direction of the field and the surface enclosed by the coil changes. As a result, the magnetic flux through the coil changes over time (the magnetic flux is given by:
 $\phi=BAcos\theta$ 
where B is the intensity of the magnetic field, A is the area enclosed by the coil and $\theta$ is the angle between the direction of B and the perpendicular to the plane of the coil). For Faraday-Newmann-Lenz law, this change in flux induces an electromotive force (emf) into the coil, according to:
 $emf=- \frac{d \Phi}{dt}$ 
where the numerator is the variation of magnetic flux and dt is the time interval. This emf in the coil produced an electrical current in the circuit.

7 0

3 years ago

Read 2 more answers

An alpha particle (the nucleus of a helium atom) consists of two protons and two neutrons, and has a mass of 6.64 * 10-27 kg. A

melamori03 [73]

Answer:

t = 4.21x10⁻⁷ s

Explanation:

The time (t) can be found using the angular velocity (ω):

$\omega = \frac{\theta}{t}$

Where θ: is the angular displacement = π (since it moves halfway through a complete circle)

We have:

$t = \frac{\theta}{\omega} = \frac{\theta}{v/r}$

Where:

v: is the tangential speed

r: is the radius

The radius can be found equaling the magnetic force with the centripetal force:

$qvB = \frac{mv^{2}}{r} \rightarrow r = \frac{mv}{qB}$

Where:

m: is the mass of the alpha particle = 6.64x10⁻²⁷ kg

q: is the charge of the alpha particle = 2*p (proton) = 2*1.6x10⁻¹⁹C

B: is the magnetic field = 0.155 T

Hence, the time is:

$t = \frac{\theta*r}{v} = \frac{\theta}{v}*\frac{mv}{qB} = \frac{\theta m}{qB} = \frac{\pi * 6.64 \cdot 10^{-27} kg}{2*1.6 \cdot 10^{-19} C*0.155 T} = 4.21 \cdot 10^{-7} s$