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vlada-n [284]
3 years ago
9

What is the relationship between atmospheric pressure and the density of gas particles in an area of decreasing pressure? (2 poi

nts)
A. As air pressure in an area decreases, the density of the gas particles in that area decreases.
B. As air pressure in an area decreases, the density of the gas particles in that area increases.
C. As air pressure in an area decreases, the density of the gas particles in that area remains constant.
D. As air pressure in an area decreases, the density of the gas particles in that area increases and decreases in an alternating pattern.

Also it's not showing earth space science, so that's why I put it as Physics.
Physics
2 answers:
Effectus [21]3 years ago
8 0

Answer:

As air pressure in an area decreases, the density of the gas particles in that area decreases.

Explanation:

ivann1987 [24]3 years ago
4 0

Answer:

A. As air pressure in an area decreases, the density of the gas particles in that area decreases.

Explanation:

At constant temperature, pressure and density of a gas are directly related.  So if pressure decreases, density decreases.

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A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas
Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

\varepsilon' =\dfrac{S_y}{E}

\varepsilon' =\dfrac{240}{110\times 1000}

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0
3 years ago
A Ferris wheel starts at rest and builds up to a final angular speed of 0.70 rad/s while rotating through an angular displacemen
PilotLPTM [1.2K]

Answer:

The average angular acceleration is 0.05 radians per square second.

Explanation:

Let suppose that Ferris wheel accelerates at constant rate, the angular acceleration as a function of change in angular position and the squared final and initial angular velocities can be clear from the following expression:

\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha\cdot (\theta-\theta_{o})

Where:

\omega_{o}, \omega - Initial and final angular velocities, measured in radians per second.

\alpha - Angular acceleration, measured in radians per square second.

\theta_{o}, \theta - Initial and final angular position, measured in radians.

Then,

\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}

Given that \omega_{o} = 0\,\frac{rad}{s}, \omega = 0.70\,\frac{rad}{s} and \theta-\theta_{o} = 4.9\,rad, the angular acceleration is:

\alpha = \frac{\left(0.70\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(4.9\,rad\right)}

\alpha = 0.05\,\frac{rad}{s^{2}}

Now, the time needed to accelerate the Ferris wheel uniformly is described by this kinematic equation:

\omega = \omega_{o} + \alpha \cdot t

Where t is the time measured in seconds.

The time is cleared and obtain after replacing every value:

t = \frac{\omega-\omega_{o}}{\alpha}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and \alpha = 0.05\,\frac{rad}{s^{2}}, the required time is:

t = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{0.05\,\frac{rad}{s^{2}} }

t = 14\,s

Average angular acceleration is obtained by dividing the difference between final and initial angular velocities by the time found in the previous step. That is:

\bar \alpha = \frac{\omega-\omega_{o}}{t}

If \omega_{o} = 0\,\frac{rad}{s},  \omega = 0.70\,\frac{rad}{s} and t = 14\,s, the average angular acceleration is:

\bar \alpha = \frac{0.70\,\frac{rad}{s} - 0\,\frac{rad}{s} }{14\,s}

\bar \alpha = 0.05\,\frac{rad}{s^{2}}

The average angular acceleration is 0.05 radians per square second.

4 0
3 years ago
Which statement best describes what energy transfer diagrams show? Energy can change form, but the total amount of energy stays
Rom4ik [11]
Energy can change form, but the total amount of energy stays the same.
3 0
3 years ago
Read 2 more answers
The substances referenced in the table are being considered for use in cooking materials. Since the materials must be melted dur
frutty [35]

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

Explanation:

The specific heat capacity of a substance indicates the amount of heat energy required to raise 1 kg of that substance by 1 degree in temperature.

Mathematically:

C=\frac{Q}{m\Delta T}

where

Q is the heat supplied to the substance

m is the mass of the substance

\Delta T is the change in temperature

Therefore, the higher the specific heat capacity of a substance, the more energy is needed to increase its temperature.

Here we can compare the specific heat capacity of the materials mentioned:

Silver: C=0.233 J/gK

Brass: C=0.380 J/gK

Platinum: C=0.130 J/gK

Aluminium: C=0.910 J/gK

Copper: C=0.390 J/gK

Therefore, the correct statements are:

Brass requires more energy than silver

Aluminum requires more energy than copper

Aluminum requires more energy than brass

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
Which of the following is an SI base unit for measuring length?
irina [24]
Meters it the SI unit for measuring length.
6 0
3 years ago
Read 2 more answers
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