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kkurt [141]
3 years ago
11

How is steam foremed

Physics
2 answers:
GenaCL600 [577]3 years ago
4 0
Steam<span> is water in the gas phase, which is </span>formed<span> when water boils. </span>Steam<span> is invisible; however, "</span>steam<span>" often refers to wet </span>steam<span>, the visible mist or aerosol of water droplets </span>formed<span> as this water vapour condenses.
hope this helps plz give brainliest!</span>
bearhunter [10]3 years ago
3 0
Steam is formed by the heating of water. When water gets too hot is forms into steam and rises into the air until it becomes cold enough to fall back down to the earth.
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A scooter has wheels with a diameter of 120 mm. What is the angular speed of the wheels when the scooter is moving forward at 6.
nirvana33 [79]

To develop this problem we will apply the concepts related to angular kinematic movement, related to linear kinematic movement. Linear velocity can be described in terms of angular velocity as shown below,

v = r\omega \rightarrow \omega = \frac{v}{r}

Here,

v = Lineal velocity

\omega= Angular velocity

r = Radius

Our values are

v = 6/ms

r = \frac{d}{2} = \frac{120*10^{-3}}{2} = 0.06m

Replacing to find the angular velocity we have,

\omega = \frac{6m/s}{0.06m}

\omega = 100rad/s

Convert the units to RPM we have that

\omega = 100rad/s (\frac{1rev}{2\pi rad})(\frac{60s}{1m})

\omega = 955.41rpm

Therefore the angular speed of the wheels when the scooter is moving forward at 6.00 m/s is 955.41rpm

4 0
3 years ago
A 2.00 kg cat is in a 97.00 kg elevator. What force on the elevator cable would be needed to lower the cat/elevator pair with an
umka21 [38]

The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.

<u>Given the following data:</u>

  • Mass of cat = 2 kg
  • Mass of elevator = 97 kg
  • Acceleration = 2 m/s^2

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:

First of all, we would calculate the total mass of the cat/elevator pair.

Total \;mass=2 + 97

Total mass = 99 kilograms

Mathematically, Newton's Second Law of Motion is given by this formula;

Force = mass \times acceleration

Substituting the given parameters into the formula, we have;

Force = 99 \times 2

Net force = 198 Newton

Read more here: brainly.com/question/24029674

3 0
2 years ago
What is the tube that carries air inside a lung?
Elis [28]
The trachea is a tube that carries air inside the lungs.
4 0
3 years ago
Read 2 more answers
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air
valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration:x_f-x_0=v_0*t+\frac{1}{2}*a*t^2, where x_f-x_0 is the distance traveled, v_0 is the initial velocity, t the time of the process and, a is the acceleration of the process.

Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

<u>Note that the acceleration is negative because is a braking process.</u>

4 0
3 years ago
Read 2 more answers
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