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3 years ago

Which statement best describes how a wave would move differently through a pot of boiling water than the steam created from it?

Physics

2 answers:

Marina86 [1]3 years ago

3 0

<span>The wave would move faster through the water than through the steam.</span>

shepuryov [24]3 years ago

3 0

Hello. This question is incomplete. The complete question is:

"Which statement best describes how a wave would move differently through a pot of boiling water than the steam created from it? The wave would move faster through the water than through the steam. The wave would move slower through the water than through the steam. The wave would move the same speed through the water and the steam. The wave would move through the water and steam at the same speed, but decrease in speed at the transition point. "

Answer:

The wave would move faster through the water than through the steam.

Explanation:

As we already know, the water molecules at the moment when the water is boiling move very fast, due to the thermal energy that acts on them. This movement causes the mileculas to collide with each other quickly, for this reason we can observe that movement of the boiling water. This makes the wave movement in the boiling water fast.

On the other hand, all this movement is not seen in water vapor molecules, since the space between them is larger and there is no energy acting directly on them. For this reason, we can say that, through steam, the wave would not move with such speed.

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Explain velocity ratio. Explain and define the efficiency of a machine

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Velocity ratio is also defined as the ratio of a distance through which any part of a machine moves, to that which the driving part moves during the same time. An object has a mechanical advantage if it exerts a force higher than the velocity ratio.

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3 years ago

Difine scalar quantity

vfiekz [6]

Scalar quantity are physical quantities that have just magnitude, not direction.

It is always positive.
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2 years ago

A 70.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 15.0 m

Levart [38]

Answer:

1.3823 rad/s

20.7345 m/s

28.66129935 m/s²

$a=2.92164g$

2006.29095 N radially outward

Explanation:

r = Radius = 15 m

m = Mass of person = 70 kg

g = Acceleration due to gravity = 9.81 m/s²

Angular velocity is given by

$\omega=13.2\times \dfrac{2\pi}{60}\\\Rightarrow \omega=1.3823\ rad/s$

Angular velocity is 1.3823 rad/s

Linear velocity is given by

$v=r\omega\\\Rightarrow v=15\times 1.3823\\\Rightarrow v=20.7345\ m/s$

The linear velocity is 20.7345 m/s

Centripetal acceleration is given by

$a_c=r\omega^2\\\Rightarrow a_c=15\times 1.3823^2\\\Rightarrow a_c=28.66129935\ m/s^2$

The centripetal acceleration is 28.66129935 m/s²

Acceleration in terms of g

$\dfrac{a}{g}=\dfrac{28.66129935}{9.81}\\\Rightarrow a=2.92164g$

$a=2.92164g$

Centripetal force is given by

$F_c=ma_c\\\Rightarrow F_c=70\times 28.66129935\\\Rightarrow F_c=2006.29095\ N$

The centripetal force is 2006.29095 N radially outward

The torque will be experienced when the centrifuge is speeding up of slowing down i.e., when it is accelerating and decelerating.

3 0

3 years ago

A 600-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 14-turn coil is wound tightly around the center of the solenoid. If

Delvig [45]

Answer:

The induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

Explanation:

The magnetic field at the center of the solenoid is given by;

B = μ(N/L)I

Where;

μ is permeability of free space

N is the number of turn

L is the length of the solenoid

I is the current in the solenoid

The rate of change of the field is given by;

$\frac{\delta B}{\delta t} = \frac{\mu N \frac{\delta i}{\delta t} }{L} \\\\\frac{\delta B}{\delta t} = \frac{4\pi *10^{-7} *600* \frac{5}{0.6} }{0.25}\\\\\frac{\delta B}{\delta t} =0.02514 \ T/s$

The induced emf in the shorter coil is calculated as;

$E = NA\frac{\delta B}{\delta t}$

where;

N is the number of turns in the shorter coil

A is the area of the shorter coil

Area of the shorter coil = πr²

The radius of the coil = 2.5cm / 2 = 1.25 cm = 0.0125 m

Area of the shorter coil = πr² = π(0.0125)² = 0.000491 m²

$E = NA\frac{\delta B}{\delta t}$

E = 14 x 0.000491 x 0.02514

E = 1.728 x 10⁻⁴ V

Therefore, the induced emf in the short coil during this time is 1.728 x 10⁻⁴ V

8 0

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AlexFokin [52]

A) as a longitudinal wave

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3 years ago