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Tresset [83]
3 years ago
6

Given the equation E = P/N , solve for P

Physics
1 answer:
Mars2501 [29]3 years ago
6 0

Answer:

E=P/N

multiply both sides by N

P=EN

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Answer:

Not collecting the info

Explanation:

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The three branches of science earth and physicals
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3 years ago
Please help me with this​
masha68 [24]

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7 0
2 years ago
Explain how the basic unit are combined to give the derived units of force, velocity, pressure and work
LuckyWell [14K]

Velocity:

Velocity is change in displacement with respect to time:

\frac{\Delta x}{\Delta t}

Analysing the units, meters (displacement) and seconds (time) are basic units:

\frac{m}{s}

Therefore the unit of velocity is m/s

Force:

Newton's second law of motion:

F = ma

Kilogram (mass) is a basic unit, and accelerations unit can be found using the equation:

a=\frac{\Delta v}{\Delta t}

Analysing the units:

\frac{\frac{m}{s}}{s}=\frac{m}{s^2}

Therefore, the unit of force is:

kg\frac{m}{s^2}

Pressure:

Pressure is given by the equation:

P=\frac{F}{S} where S is area of effect, F is force

Area for a basic rectangle (geometric shape is arbitrary for dimensional analysis) is found by multiplying two lengths:

[l^2]=m^2, the unit of area

Dividing the aforementioned unit of force by the unit of area:

\frac{kg\frac{m}{s^2}}{m^2}=\frac{kg}{ms^2}, the unit of pressure

Work:

Work is given by the equation:

W=\vec{F}\cdot \vec{x}, (dot product may be assumed as normal multiplication for the purposes of unit analysis)

Knowing displacement's (x) unit is m:

[W]=\frac{kgm}{s^2}m=\frac{kgm^2}{s^2}, the unit of work.

3 0
3 years ago
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
3 years ago
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