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4 years ago

If a gas has a gage pressure of 156 kPa, it is absolute pressure is approximately

1 answer:

5 0

In the given question, one important information for getting to the actual solution is not given and that is the atmospheric pressure. To find the approximate absolute pressure, it is needed to add the value of atmospheric pressure with the gage pressure.
Atmospheric pressure = 100 kPa
Then
Absolute pressure = 156 + 100 kPa
= 256 KPa.

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A cube, whose edges are aligned with the , and axes, has a side length . The field is immersed in an electric field aligned with

Ivenika [448]

Complete Question

Complete Question is attached below

Answer:

$q=1.558*10^{-9}c$

Explanation:

From the question we are told that:

Side length s=1.13m

Left field strength $E_l=784.75N/m$

Right field strength $E_r=776.38 N/m$

Front field strength $E_f=725.5 N/m$

Back field strength $E_b=749.54 N/m$

Top field strength $E_t=944.95 N/m$

Bottom field strength $E_{bo}=1082.58 N/m$

Generally, the equation for Charge flux is mathematically given by

$\phi=EAcos\theta$

Where

Theta for Right,Left,Front and Back are at an angle 90

$cos 90=0$

Therefore

$\phi =0$ with respect to Right,Left,Front and Back

Generally, the equation for Charge Flux is mathematically also given by

$\phi=\frac{q}{e_o}$

Where

$Area =L*B\\\\A=1.13*1.13\\\\A=1.2769m^2$

Therefore

$Q_{net}=E_{bo}Acos\theta_{bo} +E_tAcos\theta_t$

$Q_{net}=1082.85*1.2769*cos0=944.95*1.2769cos (180)$

$Q_{net}=176N/C m^2$

Giving

$q=\phi*e_0$

$q=176N/C m^2*1.558*10^{-12}c$

$q=1.558*10^{-9}c$

4 0

3 years ago

While at a party, you pull up a sound intensity level app on your phone (everyone does stuff like that, right?), and it reads 83

allochka39001 [22]

To solve this problem it is necessary to apply the concepts related to Sound Intensity. The unit most used in the logarithmic scale is the decibel and mathematically this is expressed as

$\beta_{dB} = 10log_{10}\frac{I}{I_0}$

Where,

$\beta_{dB}$ = Sound intensity level in decibels

I = Acoustic intensity on the linear scale

$I_0 =$ Hearing threshold

According to the values, the total intensity is 32 times the linear intensity and the value in decibels is 83dB

So:

$10log_{10}(\frac{32I}{I_0}) = 83$

$10log(\frac{I}{I_0})+10log(32) = 83$

$10log(\frac{I}{I_0})= 83-10log(32)$

$10log(\frac{I}{I_0})= 67.948dB$

Therefore the sound intensity due to one person is 67.948dB

8 0

3 years ago

Why is the moon’s surface cratered, but the Earth’s is not

Scorpion4ik [409]

The comets hit the moon's surface because there is no atmosphere on the moon to protect it. The earth has an atmosphere so it is protected.

8 0

3 years ago

A 50 kg aardvark runs with a speed of 6 m/s. what is the kinetic energy of the aardvark

jeyben [28]

Answer:

900 J

Explanation:

0.5 x 50= 25 x 6^2= 900 J

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4 years ago

A horizontal pull A pulls two wagons over a horizontal frictionless floor, the first wagon is 500N, the second is 2000 N. The te

MrMuchimi

<span>The force will be zero if the wagons are moving at a constant speed (i.e. not accelerating), as there is no frictional force to overcome. If the wagons are accelerating, the force will be proportional to the acceleration, and 20% of the force applied by A.</span>

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4 years ago