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schepotkina [342]
2 years ago
5

Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 2x2 − 3x

y + xyz (a) Find the rate of change of the potential at P(2, 6, 5) in the direction of the vector v = i + j − k.
Physics
1 answer:
liq [111]2 years ago
3 0

Answer:

Rate of change of potential V at the point P (2,6,5) in the direction (u = i + j - k) is

(û.∇V) = 6.928

Explanation:

V(x,y,z) = (2x² − 3xy + xyz)

Rate of change of a potential, V, in a direction, u, at a point P, is given as (û.∇V)

where û = unit vector in the direction of u = (vector u)/(magnitude of u)

u = v = i + j − k

Magnitude of u = √[(1²) + (1²) + (-1)²] = √3

û = (i + j − k)/(√3)

∇V = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)k](V) at point (2,6,5)

V(x,y,z) = (2x² − 3xy + xyz)

(∂V/∂x) = 4x - 3y + yz = 4(2) - 3(6) + (6)(5) = 20

(∂V/∂y) = - 3x + xz = -3(2) + (2)(5) = 4

(∂V/∂z) = xy = (2)(6) = 12

∇V = (20î + 4j + 12k)

Rate of change of potential V = (û.∇V)

û = (1/√3) (i + j − k)

(û.∇V) = (1/√3) [(i + j − k).(20î + 4j + 12k)]

(û.∇V) = (1/√3) [20 + 4 - 12) = 12/√3 = 6.928

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at certain times the demand for electric energy is low and electric energy is used to pump water to a reservoir 45 m above the g
Readme [11.4K]

The mass of water that must be raised is 5.25\cdot 10^7 kg

Explanation:

Since the process is 70% efficiency, the power in output to the turbine can be written as

P_{out} = 0.70 P_{in}

where P_{in} is the power in input.

The power in input can be written as

P_{in} = \frac{W}{t}

where

W is the work done in lifting the water

t = 3 h = 10,800 s is the time elapsed

The work done in lifting the water is given by

W=mgh

where

m is the mass of water

g=9.8 m/s^2 is the acceleration of gravity

h = 45 m is the height at which the water is lifted

Combining the three equations together, we get:

P_{out} = 0.70 \frac{mgh}{t}

Where

P_{out} = 150 MW = 150\cdot 10^6 W

And solving for m, we find:

m=\frac{Pt}{0.70gh}=\frac{(1.50\cdot 10^6)(10800)}{(0.70)(9.8)(45)}=5.25\cdot 10^7 kg

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

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If a negatively charged rod is held near a neutral metal ball, the ball is attracted to the rod. this happens because
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