Explain the differences among sand, silt, and clay, both in their physical characteristics and their behavior in relation to bui
1 answer:
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a) Potential in A: -2700 V
b) Potential difference: -26,800 V
c) Work:
Explanation:
a)
The electric potential at a distance r from a single-point charge is given by:
where
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
and is located at the origin (x=0, y=0)
Charge 2 is
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
So the potential due to charge 2 is
The distance of point A from charge 2 is
So the potential due to charge 1 is
Therefore, the net potential at point A is
b)
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
So the potential due to charge 1 at point B is
The distance of charge 2 from point B is
So the potential due to charge 2 at point B is
Therefore, the net potential at point B is
So the potential difference is
c)
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
where
q is the charge of the particle
is the potential difference
In this problem, we have:
is the charge of the electron
is the potential difference
Therefore, the work required on the electron is
Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer: The exit temperature of the gas in deg C is .
Explanation:
The given data is as follows.
= 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)
= 100 kPa,
We know that for an ideal gas the mass flow rate will be calculated as follows.
or, m =
=
= 10 kg/s
Now, according to the steady flow energy equation:
= 5 K
= 5 K + 300 K
= 305 K
= (305 K - 273 K)
=
Therefore, we can conclude that the exit temperature of the gas in deg C is .