different between an architect and an engineer
Civil engineers work on the construction of the building. ... In short it can be said that while an Architect is a building designer, a Civil Engineer is a structural expert, who in
Answer:
Resistance of copper = 1.54 * 10^18 Ohms
Explanation:
<u>Given the following data;</u>
Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m
Resistivity, P = 1.7 * 10^8 Ωm
Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m
Radius = 0.000325 m
To find the resistance;
Mathematically, resistance is given by the formula;
![Resistance = P \frac {L}{A}](https://tex.z-dn.net/?f=%20Resistance%20%3D%20P%20%5Cfrac%20%7BL%7D%7BA%7D%20)
Where;
- P is the resistivity of the material.
- L is the length of the material.
- A is the cross-sectional area of the material.
First of all, we would find the cross-sectional area of copper.
Area of circle = πr²
Substituting into the equation, we have;
Area = 3.142 * (0.000325)²
Area = 3.142 * 1.05625 × 10^-7
Area = 3.32 × 10^-7 m²
Now, to find the resistance of copper;
![Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}}](https://tex.z-dn.net/?f=%20Resistance%20%3D%201.7%20%2A%2010%5E%7B8%7D%20%5Cfrac%20%7B3000%7D%7B3.32%20%2A%2010%5E%7B-7%7D%7D%20)
![Resistance = 1.7 * 10^{8} * 903614.46](https://tex.z-dn.net/?f=%20Resistance%20%3D%201.7%20%2A%2010%5E%7B8%7D%20%2A%20903614.46%20)
<em>Resistance = 1.54 * 10^18 Ohms </em>
Answer:
5
Explanation:
3/4of marble is blue
i.e,3/4 of 20
i.e, 15 are blue
then number of NOT blue is 20-15=5!
✌️
Answer:
a) The difference in mercury levels in the manometer is 2 centimeters.
b) The gage of the gas is 2.670 kilopascals.
Explanation:
a) Pressure in gases is absolute. A manometer helps to determine the hydrostatic difference between pressure of the gas (
) and atmospheric pressure (
), both measured in pascals. A kilopascal equals 1000 pascals and 1 meter equals 100 centimeters. That is:
(1)
Where:
- Density of mercury, measured in kilograms per cubic meter.
- Gravitational acceleration, measured in meters per square second.
- Difference in mercury levels, measured in meters.
If we know that
,
,
and
, the difference in mercury levels in the manometer is:
![L = \frac{P_{g}-P_{atm}}{\rho\cdot g}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7BP_%7Bg%7D-P_%7Batm%7D%7D%7B%5Crho%5Ccdot%20g%7D)
![L = \frac{104000\,Pa-101330\,Pa}{\left(13590\,\frac{kg}{m^{3}} \right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7B104000%5C%2CPa-101330%5C%2CPa%7D%7B%5Cleft%2813590%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%5Cright%29%5Ccdot%20%5Cleft%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%7D)
![L = 0.020\,m](https://tex.z-dn.net/?f=L%20%3D%200.020%5C%2Cm)
![L = 2\,cm](https://tex.z-dn.net/?f=L%20%3D%202%5C%2Ccm)
The difference in mercury levels in the manometer is 2 centimeters.
b) The gage pressure is the difference between gas pressure and atmospheric pressure: (
,
)
(2)
![P_{gage} = 104000\,Pa-101330\,Pa](https://tex.z-dn.net/?f=P_%7Bgage%7D%20%3D%20104000%5C%2CPa-101330%5C%2CPa)
![P_{gage} = 2670\,Pa](https://tex.z-dn.net/?f=P_%7Bgage%7D%20%3D%202670%5C%2CPa)
![P_{gage} = 2.670\,kPa](https://tex.z-dn.net/?f=P_%7Bgage%7D%20%3D%202.670%5C%2CkPa)
The gage of the gas is 2.670 kilopascals.
Answer:
![\Delta V_{gas}=0.018m^3](https://tex.z-dn.net/?f=%5CDelta%20V_%7Bgas%7D%3D0.018m%5E3)
Explanation:
We define as,
c.V Piston
![(E_2-E_1)_{Pist}=m(u_2-u_1)+m[\frac{1}{2}V^2_2-0]+mg(h_2-0)](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3Dm%28u_2-u_1%29%2Bm%5B%5Cfrac%7B1%7D%7B2%7DV%5E2_2-0%5D%2Bmg%28h_2-0%29)
![(E_2-E_1)_{Pist}=0+25*0.5*25^2+25*9.8*5](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3D0%2B25%2A0.5%2A25%5E2%2B25%2A9.8%2A5)
![(E_2-E_1)_{Pist}=7712+5+1225.8=9038.3J](https://tex.z-dn.net/?f=%28E_2-E_1%29_%7BPist%7D%3D7712%2B5%2B1225.8%3D9038.3J)
The energy equation for the piston is
![E_2-E_1=W_{gas}-W_{atm}=P_{avg}\Delta V_{gas}-P_0 \Delta V_{gas}](https://tex.z-dn.net/?f=E_2-E_1%3DW_%7Bgas%7D-W_%7Batm%7D%3DP_%7Bavg%7D%5CDelta%20V_%7Bgas%7D-P_0%20%5CDelta%20V_%7Bgas%7D)
Remember that
![\Delta V_{atm}=-\Delta V_{gas}](https://tex.z-dn.net/?f=%5CDelta%20V_%7Batm%7D%3D-%5CDelta%20V_%7Bgas%7D)
So,
![\Delta V_{gas}=9.038kJ/(600-100)=0.018m^3](https://tex.z-dn.net/?f=%5CDelta%20V_%7Bgas%7D%3D9.038kJ%2F%28600-100%29%3D0.018m%5E3)