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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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mrs_skeptik [129]

Answer:

duty h gucuuvu h just hc i oicuxp o cut o icucj x uc jo 8cuc8c

5 0
2 years ago
A mass m neutron has elastic collision with a mass m'
hoa [83]

Answer:

The neutron loses all of its kinetic energy to nucleus.

Explanation:

Given:

Mass of neutron is 'm' and mass of nucleus is 'm'.

The type of collision is elastic collision.

In elastic collision, there is no loss in kinetic energy of the system. So, total kinetic energy is conserved. Also, the total momentum of the system is conserved.

Here, the nucleus is still. So, its initial kinetic energy is 0. So, the total initial kinetic energy will be equal to kinetic energy of the neutron only.

Now, final kinetic energy of the system will be equal to the initial kinetic energy.

Now, as the nucleus was at rest initially, so the final kinetic energy of the nucleus will be equal to the initial kinetic energy of the neutron.

Thus, all the kinetic energy of the neutron will be transferred to the nucleus and the neutron will come to rest after collision.

Therefore, the neutron loses all of its kinetic energy to nucleus.

5 0
3 years ago
Each blade of a fan has a radius of 11 inches. If the fan’s rate of turn is 1440o /sec, find the following. (a) The angular spee
notka56 [123]

Answer:

a) 24.43 radians per second

b) 268.73 inches per second

Explanation:

a) The angular speed of the fan on Celsius degrees/second is 1400, so we should convert that value to radians using the fact that 2π rad = 360 °C:

\omega = 1400\frac{C}{s}=1400\frac{C}{s}*\frac{2\pi\,rad}{360\,C}

\omega = 1400\frac{C}{s}=24.43\frac{rad}{s}

b) Linear speed on a point of the blade is related with angular speed of the fan by the equation

v=\omega r

with v linear speed, ω angular speed and r the radius of the blades. So:

v=(24.43\frac{rad}{s})(11 in)

Radians isn't really a unity; it is dimensionless so we can put it or not. So:

v=268.73\frac{in}{s}

3 0
3 years ago
What is the force exerted on a charge of 2. 5 µC moving perpendicular through a magnetic field of 3. 0 × 102 T with a velocity o
stira [4]

The force acting on a moving charge is known as the magnetic force. The force acting on the charge will be 3.75 N.

<h3>What is the force exerted on the charge?</h3>

Magnetic fields only exert a force on a moving electric charge. A moving charge generates a magnetic field. With an increase in charge and magnetic field strength, this force rises.

when charges have higher velocities, the force is stronger. However, the magnetic force is always perpendicular to the velocity.

Mathematically the force exerted on the charge will be

F=qvBsinα

F= force acting on the charge

v = velocity of charge

q = charge

F=qvBsinα

F=2.5×10⁻⁶×5.0×10³×3.0×10²

F=37.5 N

Hence The force acting on the charge will be 3.75 N.

To learn more about the force acting on charge refer to ;

brainly.com/question/451411

F = q V B sinα

Where F is the force applied to a moving charge.

V = charge velocity

q stands for charge.

α = angle between V and B directions

As a result, the moving charge is subjected to a force of 3.75 Newton.

3 0
1 year ago
Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and
tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

&

q_x = -k*(dT/dx)

<u>Case (1)  </u>

dT/dx= (-20-50)/0.35==> -280 K/m

 q_x  =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

 q_x  =-50*(80)*10^3==>-4 kW

Case (3)

q_x  =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x  =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x  =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0
3 years ago
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