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irinina [24]
3 years ago
13

What material parameters determine resistivity?

Physics
1 answer:
sleet_krkn [62]3 years ago
3 0

Answer:

Resistivity \rho =\frac{RA}{l}

It depends upon cross sectional area and length of material

Explanation:

The resistance of any material is given by R=\frac{\rho l}{A}, here \rho is the resistivity of material , l is length of material and A is cross sectional area

So resistivity \rho =\frac{RA}{l}

So resistuivity of any material depends upon area of cross section and length of material

If cross sectional area will be more then resistivity will be more. And is length of the material will be more then resistivity will be less

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Answer:

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Explanation:

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6 0
2 years ago
How do you test with crash dummies seat belts and airbags Illustrate Newton’s first law of motion
Harrizon [31]
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3 0
2 years ago
A plane progressive wave is represented by the equation y =2sin (2000πt- 0.5x) the symbols have their usual meanings. what is th
frutty [35]

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5 0
3 years ago
Steam undergoes an adiabatic expansion in a piston–cylinder assembly from 100 bar, 360°C to 1 bar, 160°C. What is work in kJ per
vfiekz [6]

Answer:

work is 130.5 kJ/kg

entropy change is 1.655 kJ/kg-k

maximum  theoretical work is 689.4 kJ/kg

Explanation:

piston cylinder assembly

100 bar, 360°C to 1 bar, 160°C

to find out

work  and amount of entropy  and magnitude

solution

first we calculate work i.e heat transfer - work =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C    .................1

so first we get some value from steam table with the help of 100 bar @360°C and  1 bar @ 160°C

specific volume = 0.0233 m³/kg

specific enthalpy = 2961 kJ/kg

specific internal energy = 2728 kJ/kg

specific entropy = 6.004 kJ/kg-k

and respectively

specific volume = 1.9838 m³/kg

specific enthalpy = 2795.8 kJ/kg

specific internal energy = 2597.5 kJ/kg

specific entropy = 7.659 kJ/kg-k

now from equation 1 we know heat transfer q = 0

so - w =   specific internal energy @1 bar, 160°C  - specific internal energy @ 100 bar, 360°C

work = 2728 - 2597.5

work is 130.5 kJ/kg

and entropy change formula is i.e.

entropy change =  specific entropy ( 100 bar @360°C)  - specific entropy ( 1 bar @160°C )

put these value we get

entropy change =  7.659 - 6.004

entropy change is 1.655 kJ/kg-k

and we know maximum  theoretical work = isentropic work

from steam table we know specific internal energy is 2038.3 kJ/kg

maximum  theoretical work = specific internal energy - 2038.3

maximum  theoretical work = 2728 - 2038.3

maximum  theoretical work is 689.4 kJ/kg

3 0
3 years ago
Approximate the field by treating the disk as a +6.1 C point charge at a distance of 21 cm Answer in units of N/C.
DaniilM [7]

Answer:

Electric field, E=1.24\times 10^{12}\ N/C

Explanation:

It is given that,

Charge, Q = +6.1 C

Distance, r = 21 cm = 0.21 m

We need to find the electric field. It is given by :

E=k\dfrac{Q}{r^2}

E=9\times 10^9\times \dfrac{6.1}{(0.21)^2}

E=1.24\times 10^{12}\ N/C

So, the electric field at this distance is 1.24\times 10^{12}\ N/C. Hence, this is the required solution.

7 0
3 years ago
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