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Helga
16 days 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 [2.4K]16 days 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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kicyunya [2264]

Answer:

\Delta T = 0.81 ^oC

Explanation:

According to the principle of energy conservation

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\frac{1}{2}mv^2 = ms\Delta T

Next, divide both sides by the object's mass

\frac{1}{2}v^2 = s\Delta T

the resulting temperature change is expressed as

\Delta T = \frac{v^2}{2s}

\Delta T = \frac{25^2}{2\times 387}

\Delta T = 0.81^oC

3 0
12 days ago
A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
Ostrovityanka [2204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

m_1u_1+m_2u_2=(m_1+m_2)v

v=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}

v=\dfrac{0.05\ kg\times 12\ m/s+0}{0.05\ kg+0.15\ kg}

The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

3 0
19 days ago
Albert uses as his unit of length (for walking to visit his neighbors or plowing his fields) the albert (a), the distance albert
Yuliya22 [2420]

To tackle this question, we know the following:

1 Albert equals 88 meters.

1 A = 88 m.

Initially, we square both sides of the equation:

(1 A)^2 = (88 m)^2

1 A^2 = 7,744 m^2

<span>Since 1 acre equals 4,050 m^2, let’s divide both sides by 7,744 to find out how many acres match this value:</span>

1 A^2 / 7,744 = 7,744 m^2 / 7,744

(1 / 7,744) A^2 = 1 m^2

Then multiply both sides by 4,050.

(4050 / 7744) A^2 = 4050 m^2

0.523 A^2 = 4050 m^2

<span>Thus, one acre is approximately 0.52 square alberts.</span>

7 0
24 days ago
Suppose a new asteroid was recently discovered which takes 557 months to orbit the Sun once (that's equal to 16,700 days or 46.4
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Answer:

The average distance of the new asteroid from the Sun is estimated to be (2.02 × 10⁶) km.

Explanation:

The orbital speed of planets varies based on their distance from the Sun, which also affects their orbital period.

With its 557 months, equivalent to 46.4 years for an orbit around the Sun, the new asteroid's speed is situated between the orbital speeds of Saturn and Uranus.

Uranus orbits the Sun in 84 years at 24.61 km/hour,

while Saturn completes its orbit in 29.4 years at 34.82 km/hour.

To interpolate the speed for our asteroid at 46.4 years,

we denote its speed as x.

84 years ----> 24.61 km/h

46.4 years ----> x km/h

29.4 years -----> 34.82 km/h

Setting up the proportion:

(84 - 46.4)/(46.4 - 29.4) = (24.61 - x)/(x - 34.82)

Solving for x gives the asteroid's speed as 31.64 km/hr.

To find the average speed, use the formula:

Average speed = (total distance)/(time taken).

The total distance covered equals the circumference of the orbit around the Sun = 2πR,

where R = distance from the asteroid to the Sun.

Time taken = 16700 days = 16700 × 24 hours = 400800 hours.

Thus, we find that 31.64 = (2πR)/400800.

From this, we get 2πR = 31.64 × 400800 = 12681312 km.

And, R = 12681312/(2π) = 2018293.5 km = (2.02 × 10⁶) km.

8 0
18 days ago
When you apply the torque equation ∑τ = 0 to an object in equilibrium, the axis about which torques are calculated:
Softa [2029]

Answer:

option D.

Explanation:

The correct choice is option D.

For an object in equilibrium, the torque measured at any point will be zero.

An object is deemed to be in equilibrium when the net moment acting on it equals zero.

If the object experiences a net moment not equal to zero, it will rotate and will not remain stable.

3 0
1 month ago
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