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A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p
1 answer:
Response:
Clarification:
We need an expression that shows how much water has been drained from the tub. This is represented by v, which indicates how many gallons have flowed out since the plug was taken out. Each gallon removed equates to 8.345 pounds of water, so the weight of the drained water Q in pounds as a function of v can be expressed as:
Where v signifies the number of gallons emptied from the tub.
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Answer:
≈ 3077.34
Explanation:
To calculate the flux of F (vector field) across surface S, where
F(x,y,z) = y i − x j +and S(u,v) = u cos v i + u sin v j + v k, 0 ≤ u ≤ 2, 0 ≤ v ≤ 5π
We will evaluate the following integral:
Substituting for the surface
x = u cos v
y = u sin v
z = v
Then
F(S(u,v)) = u sin v i - u cos v j + k
The normal vector N is computed as
Where:
N = < cos v, sin v, 0 > X <- u sin v, u cos v, 2v
N = < 2v sin v, -2v cos v, u >
F(S(u,v)).N = < u sin v, -u cos v, >. < 2v sin v, -2v cos v, u >
F(S(u,v)).N = 2uv + u
Thus
≈ 3077.34
Response:
E = ρ ( R1²) / 2 ∈o R
Clarification:
Provided information
Two cylinders are aligned parallel
Distance = d
Radial distance = R
d < (R2−R1)
To determine
Express the response using the variables ρE, R1, R2, R3, d, R, and constants
Solution
We have two parallel cylinders
therefore, area equals 2 R × l
And we apply Gauss's Law
EA = Q(enclosed) / ∈o......1
Initially, we calculate Q(enclosed) = ρ Volume
Q(enclosed) = ρ ( R1² × l )
Thus, inserting all values into equation 1
produces
EA = Q(enclosed) / ∈o
E(2 R × l) = ρ (
R1² × l ) / ∈o
This simplifies to
E = ρ ( R1²) / 2 ∈o R