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Blizzard [7]
3 years ago
9

he measured flow rate of water through a 20 mm diameter pipe is 75 L/min. What is the average velocity (in m/s) of water flowing

through the pipe
Physics
1 answer:
Masteriza [31]3 years ago
6 0

Answer:

Average velocity of water flow through the pipe = 3.98 m/s

Explanation:

Diameter of pipe = 20 mm

Discharge = 75 L/min

We know that

        Discharge = Area x Velocity

      \texttt{Area of pipe}=\frac{\pi d^2}{4}=\frac{\pi \times (20\times 10^{-3})^2}{4}=3.14\times 10^{-4}m^2

       Discharge = 75 L/min

                         =\frac{75\times 10^{-3}}{60}=1.25\times 10^{-3}m^3/s

Substituting

        1.25\times 10^{-3}=3.14\times 10^{-4}\times V\\\\V=3.98m/s

Average velocity of water flow through the pipe = 3.98 m/s

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A soccer ball of mass 0.4 kg is moving horizontally with a speed of 20 m/s when it is kicked by a player. The kicking force is s
amid [387]

Answer:

v_{f} = 74 m/s, F = 230 N

Explanation:

We can work on this exercise using the relationship between momentum and moment

        I = ∫ F dt = Δp

bold indicates vectors

we can write this equations in its components

X axis

       Fₓ t = m ( -v_{xo})

Y axis  

        t = m (v_{yf} - v_{yo})

in this case with the ball it travels horizontally v_{yo} = 0

Let's use trigonometry to write the final velocities and the force

        sin 30 = v_{yf} / vf

        cos 30 = v_{xf} / vf

        v_{yf} = vf sin 30

        v_{xf} = vf cos 30

         sin40 = F_{y} / F

         F_{y} = F sin 40

         cos 40 = Fₓ / F

         Fₓ = F cos 40

let's substitute

      F cos 40 t = m ( cos 30 - vₓ₀)

      F sin 40 t = m (v_{f} sin 30-0)

we have two equations and two unknowns, so the system can be solved

        F cos 40 0.1 = 0.4 (v_{f} cos 30 - 20)

        F sin 40 0.1 = 0.4 v_{f} sin 30

we clear fen the second equation and subtitles in the first

         F = 4 sin30 /sin40     v_{f}

         F = 3.111 v_{f}

        (3,111 v_{f}) cos 40 = 4 v_{f} cos 30 - 80

        v_{f} (3,111 cos 40 -4 cos30) = - 80

        v_{f} (- 1.0812) = - 80

        v_{f} = 73.99

        v_{f} = 74 m/s

now we can calculate the force

          F = 3.111 73.99

          F = 230 N

6 0
3 years ago
How are calculation for velocity and speed different
BARSIC [14]

Only velocity uses direction of travel in its calculations.

5 0
3 years ago
You hear the engine roaring on a race car at the starting line. Predict the changes in the sound as the race starts and the car
gogolik [260]

B. The sound of the engine will get louder and the pitch higher.

4 0
3 years ago
Read 2 more answers
A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l
Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>
  • Consider a body of mass m kept on a weighing machine in a lift.
  • The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
  • The reaction we get as the weight recorded by the machine, and it is called the apparent weight.
<h3>How to solve the question?</h3>
  • Here we have given with the actual weight of the body as 100lbs.
  • This 100lb child is standing on the scale or the weighing machine, when it is riding .
  • During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
  • There is also<em> mg </em>downwards and a normal reaction in the upward direction.
  • when we equate both the upward force and downward force, we get,

                             ma=mg-N\\N=mg-ma    i.e. during riding the scale reads a weight less than that of actual weight.

  • When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

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7 0
2 years ago
The magnetic field at the center of a 1.40-cm-diameter loop is 2.50 mT . PART A) What is the current in the loop?
NISA [10]

Explanation:

It is given that,

Diameter of loop, d = 1.4 cm

Radius of loop, r = 0.7 cm = 0.007 m

Magnetic field, B=2.5\ mT=2.5\times 10^{-3}\ T

(A) Magnetic field of a current loop is given by :

B=\dfrac{\mu_oI}{2r}

I is the current in the loop

I=\dfrac{2Br}{\mu_o}

I=\dfrac{2\times 2.5\times 10^{-3}\times 0.007}{4\pi \times 10^{-7}}

I = 27.85 A

(B) Magnetic field at a distance r from a wire is given by :

B=\dfrac{\mu_o I}{2\pi r}

r=\dfrac{\mu_o I}{2\pi B}

r=\dfrac{4\pi \times 10^{-7}\times 27.85}{2\pi \times 2.5\times 10^{-3}}

r = 0.00222 m

r=2.2\times 10^{-3}\ m

Hence, this is the required solution.

3 0
3 years ago
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