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3 years ago

Is there a way to get the answers to a NCCER book test?

Engineering

1 answer:

sergeinik [125]3 years ago

3 0

Answer:

go on google and type NCEER book answers

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A furniture manufacturer purchases a drill press machine enabled with 5G and edge computing capabilities to keep the machine ope

Andreas93 [3]

A lot of manufacturer often uses 5G machines. How these capabilities could help improve safety of the operators is that it does includes an emergency switch for the operator so that one can manually shut off when needed.

<h3>Edge computing with 5G</h3>

The edge computing along with 5G network and IoT devices can help put together different safety features and limitations and on can use them to known the unsafe action and also data can be communicated.

Edge computing when use with 5G produces good opportunities in all industry. It is known to help bring computation and data storage close to where data is been produced and it enable good data control, reduced costs, etc.

Learn more about 5G network from

brainly.com/question/24664177

8 0

2 years ago

For a certain gas, Cp = 840.4 J/kg-K; and Cv = 651.5 J/kg-K. How fast will sound travel in this gas if it is at an adiabatic sta

Crank

Answer:

The speed of the sound for the adiabatic gas is 313 m/s

Explanation:

For adiabatic state gas, the speed of the sound c is calculated by the following expression:

$c=\sqrt(\gamma*R*T)$

Where R is the gas's particular constant defined in terms of Cp and Cv:

$R=Cp-Cv$

For particular values given:

$R=840.4 \frac{J}{Kg-K}- 651.5 \frac{J}{Kg-K}$

$R=188.9 \frac{J}{Kg-K}$

The gamma undimensional constant is also expressed as a function of Cv and Cp:

$\gamma=Cp/Cv$

$\gamma=840.4 \frac{J}{Kg-K} / 651.5 \frac{J}{Kg-K}$

$\gamma=1.29$

And the variable T is the temperature in Kelvin. Thus for the known temperature:

$c=\sqrt(1.29*188.9 \frac{J}{Kg-K}*377 K)$

$c=\sqrt(91867.73 \frac{J}{Kg})$

The Jules unit can expressing by:

$J=N.m=\frac{Kg.m}{s^2}* m$

$J=\frac{Kg.m^2}{s^2}$

Replacing the new units for the speed of the sound:

$c=\sqrt(91867.73 \frac{Kg.m^2}{Kg.s^2})$

$c=\sqrt(91867.73 \frac{m^2}{s^2})$

$c=313 m/s$

3 0

3 years ago

Read 2 more answers

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

6 0

1 year ago

A charge of +2.00 μC is at the origin and a charge of –3.00 μC is on the y axis at y = 40.0 cm . (a) What is the potential at po

Nimfa-mama [501]

a) Potential in A: -2700 V

b) Potential difference: -26,800 V

c) Work: $4.3\cdot 10^{-15} J$

Explanation:

The electric potential at a distance r from a single-point charge is given by:

$V(r)=\frac{kq}{r}$

where

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

q is the charge

r is the distance from the charge

In this problem, we have a system of two charges, so the total potential at a certain point will be given by the algebraic sum of the two potentials.

Charge 1 is

$q_1=+2.00\mu C=+2.00\cdot 10^{-6}C$

and is located at the origin (x=0, y=0)

Charge 2 is

$q_2=-3.00 \mu C=-3.00\cdot 10^{-6}C$

and is located at (x=0, y = 0.40 m)

Point A is located at (x = 0.40 m, y = 0)

The distance of point A from charge 1 is

$r_{1A}=0.40 m$

So the potential due to charge 2 is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.40}=+4.50\cdot 10^4 V$

The distance of point A from charge 2 is

$r_{2A}=\sqrt{0.40^2+0.40^2}=0.566 m$

So the potential due to charge 1 is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.566}=-4.77\cdot 10^4 V$

Therefore, the net potential at point A is

$V_A=V_1+V_2=+4.50\cdot 10^4 - 4.77\cdot 10^4=-2700 V$

Here we have to calculate the net potential at point B, located at

(x = 0.40 m, y = 0.30 m)

The distance of charge 1 from point B is

$r_{1B}=\sqrt{(0.40)^2+(0.30)^2}=0.50 m$

So the potential due to charge 1 at point B is

$V_1=\frac{(8.99\cdot 10^9)(+2.00\cdot 10^{-6})}{0.50}=+3.60\cdot 10^4 V$

The distance of charge 2 from point B is

$r_{2B}=\sqrt{(0.40)^2+(0.40-0.30)^2}=0.412 m$

So the potential due to charge 2 at point B is

$V_2=\frac{(8.99\cdot 10^9)(-3.00\cdot 10^{-6})}{0.412}=-6.55\cdot 10^4 V$

Therefore, the net potential at point B is

$V_B=V_1+V_2=+3.60\cdot 10^4 -6.55\cdot 10^4 = -29,500 V$

So the potential difference is

$V_B-V_A=-29,500 V-(-2700 V)=-26,800 V$

The work required to move a charged particle across a potential difference is equal to its change of electric potential energy, and it is given by

$W=q\Delta V$

where

q is the charge of the particle

$\Delta V$ is the potential difference

In this problem, we have:

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=-26,800 V$ is the potential difference

Therefore, the work required on the electron is

$W=(-1.6\cdot 10^{-19})(-26,800)=4.3\cdot 10^{-15} J$

4 0

3 years ago

A discrete MOSFET common-source amplifier has RG = 2 MΩ, gm = 5 mA/V, ro = 100 kΩ, RD = 20kΩ, Cgs = 3pF, and Cgd = 0.5pF. The am

Papessa [141]

Answer:

a) -36.36 V/V

b) 15.17 kHz

c) 1.6 GHz

Explanation:

See attached picture.

7 0

3 years ago