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Helga
3 months ago
12

When you urinate, you increase pressure in your bladder to produce the flow. For an elephant, gravity does the work. An elephant

urinates at a remarkable rate of 0.0060 m3 (a bit over a gallon and a half) per second. Assume that the urine exits 1.0 m below the bladder and passes through the urethra, which we can model as a tube of diameter 8.0 cm and length 1.2 m. Assume that urine has the same density as water, and that viscosity can be ignored for this flow.
1) What is the speed of the flow?
2) If we assume that the liquid is at rest in the bladder (a reasonable assumption) and that the pressure where the urine exits is equal to atmospheric pressure, what does Bernoulli's equation give for the pressure in the bladder?
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
8 0

Answer:

a) v = 1.19 m/s, b) P₁ = 0.922 x 10⁵ Pa

Explanation:

1) We will apply the fluid continuity equation

       Q = A v

The area of a circle is

      A = π r² = π d²/4

     

     v = Q / A = Q 4 / π d²

     v = 0.006 4/π 0.08²

     v =  1.19 m/s

2) Apply Bernoulli's equation, considering point 1 as the bladder and point 2 as where the urine exits

     P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

The problem states that

P₂ = 1.0013 x 10⁵ Pa

v₁ = 0

y₁ = 1 m

y₂ = 0

Density of water (ρ) = 1000 kg/m³

      P₁ + ρ y₁ = P₂ + ½ ρ v₂²

      P₁ = P₂ + ½ ρ v₂² - ρ g y₁

      P₁ = 1.013 x 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1

      P₁ = 1.013 x 10⁵ + 708.5  - 9800

      P₁ =  92208.5 Pa

      P₁ = 0.922 x 10⁵ Pa

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Answer and Explanation:

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According to Ampere's law

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3 months ago
. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it
Keith_Richards [3271]

Ans    specifically, 4

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7 0
3 months ago
Part A
kicyunya [3294]

Answer:

v' = -18 m/s

Explanation:

  • Assuming no external forces act during the collision, the overall momentum must be preserved as follows:

       p_{o} = p_{f} (1)

  • The initial momentum can be defined this way (with the initial direction of the ball considered positive):

       m_{b} * v_{b} -M_{c}*V_{c} = m_{b} * 18 m/s + (-M_{c}* 20 m/s) (2)

  • The final momentum can be formulated as follows (given that v'b moves in the opposite direction to vb):

        -(m_{b} * v'_{b}) + M_{c}*V'_{c} (3)

  • If we consider that Mc >> mb, we assume the car's speed remains unchanged as a result of the collision, allowing us to substitute V'c for Vc in (3).
  • Consequently, we can rewrite (3) like this:

       -(m_{b} * v'_{b}) +(- M_{c}*V_{c}) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (4)

  • By substituting (2) and (4) back into (1), we arrive at:

       m_{b} * 18 m/s + (-M_{c}* 20 m/s) = -(m_{b} * v'_{b}) + (-M_{c} * 20 m/s) (5)

  • By simplifying and rearranging, we can solve for v'b like this:
  • v'_{b} = -18 m/s (6), which makes sense because it's as though the ball collided with a wall and simply reversed its velocity post-collision.
3 0
3 months ago
Which formulas show the relationships between momentum, mass, and velocity? Check all that apply.
serg [3582]
A straightforward way to visualize this is through the relationship in a triangle: P
M V
where p represents momentum
m refers to mass
and v stands for velocity
The formulas are:
p = m x v
m = p/v
v = p/m
Thus, all provided options are correct
4 0
3 months ago
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