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pantera1 [17]
2 years ago
12

An incompressible fluid flows at .252 m/s through a 44 diameter (circular cross section) pipe. The pipe widens to a square cross

sectional area 5.5 cm on a side. Assuming steady flow:
A. What is the speed through the square section

B. What is the flow rate in liters/minute

Work/Process is appreciated!
Physics
1 answer:
Serggg [28]2 years ago
7 0

ANSWER:

The easiest way to get a fairly accurate measure of your water flow rate is to time yourself filling up a bucket. So for example if you fill up a 10 litre bucket in 1.5 minutes, then your flow rate will be: 10/1.5 = 6.66 Litres per minute.

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A car moving with an intial velocity of 60m/s is brought to rest in 30 seconds calculate the acceleration
gregori [183]

Answer:

a = 2 [m/s^2]

Explanation:

To solve this problem we must use the expressions of kinematics, we must bear in mind that when a body is at rest its velocity is zero.

v_{f} = v_{i} - (a*t)

where:

Vf = final velocity = 0

Vi = initial velocity = 60 [m/s]

a = desacceleration [m/s^2]

t = time = 30 [s]

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When the speed and direction of an object remain the same, what happens to its velocity?
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2 years ago
A landing craft with mass 1.21×104 kg is in a circular orbit a distance 5.90×105 m above the surface of a planet. The period of
creativ13 [48]

Answer:

  W = 661.6 N

Explanation:

The weight of a body is the force of attraction of the plant on the body, so we must use the law of gravitational attraction

       F = G m M / r²

Where G is the gravitational attraction constant that values ​​6.67 10-11 N m² / kg², M is the mass of the planet and r is the distance from the center of the planet.

Let's look for the mass of the planet, for this we write Newton's second law for the landing craft

     F = m a

Acceleration is centripetal a = v² / r

     G m M / r² = m (v² / r)

The ship rotates rapidly (constant velocity module), let's use uniform kinematic relationships

    v = d / t

The distance of a circle is

    d = 2π r

    v = 2π r / t

We replace

     G m M / r² = m (4π² r² / t² r)

    G M = 4 π² r³ / t²

    M = 4π² r³ / G t²

The measured distance r from the center of the plant is

     r = R orbit + R planet

     r = 5.90 10⁵ + ½ 9.80 10⁶

     r = 5.49 10⁶ m

    M = 4 π² (5.49 10⁶)³ / (6.67 10⁻¹¹ (5.900 10³)²)

    M = 6,532 10²¹ / 2,321 10⁺³

    M = 2.814 10²⁴ kg

With this data we calculate the astronaut's weight

     W = (G M / R²) m

     W = (6.67 10⁻¹¹ 2,816 10²⁴ /(4.90 10⁶)2)   84.6

     W = 7.82  84.6

     W = 661.57 N

3 0
3 years ago
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