Affordability is most concerned with:

A 11-cm-diameter horizontal jet of water with a velocity of 40 m/s relative to the ground strikes a flat plate that is moving in

Reika [66]

Answer:

F = 8552.7N

Explanation:

We need first our values, that are,

$V_{jet} = 40m/s\\V_{Plate} = 10m/s \\D = 11cm$

We start to calculate the relative velocity, that is,

$V_r = V_{jet}-V_{plate}\\V_r = (40)-(10)\\V_r = 30m/s$

With the relative velocity we can calculate the mass flow rate, given by,

$\dot{m}_r = \rho A V_r$

$\dot{m}_r = (1000)(30) \frac{\pi (0.11)^2}{4}$

$\dot{m}_r = 285.09kg/s$

We need to define the Force in the direction of the flow,

$\sum\vec{F} = \sum_{out} \beta\dot{m}\vec{V} - \sum_{in} \beta\dot{m} \vec{V}\\$

$F = \dot{m}V_r$

$F = (285.09Kg/s)(30)$

$F = 8552.7N$

8 0

3 years ago

Select the correct answer.

olya-2409 [2.1K]

Answer:

screw is the answer of the question

6 0

3 years ago

Read 2 more answers

List one advantage and one disadvantage of the use of the commutator?

Mariulka [41]

Answer and Explanation:

Commutator are used in DC machine commutator is mainly used for the reversing the direction of the current .It is connected to the armature of the DC generator or motor

ADVANTAGE OF COMMUTATOR The main advantage of the commutator in DC motor is to keep keep the direction of the toque always in the same direction by changing the current direction

DISADVANTAGE OF COMMUTATOR : The main disadvantage is due to the friction between the commutator and brushes there is a friction loss.

3 0

3 years ago

Water at atmospheric pressure boils on the surface of a large horizontal copper tube. The heat flux is 90% of the critical value

masya89 [10]

Answer:

The tube surface temperature immediately after installation is 120.4°C and after prolonged service is 110.8°C

Explanation:

The properties of water at 100°C and 1 atm are:

pL = 957.9 kg/m³

pV = 0.596 kg/m³

ΔHL = 2257 kJ/kg

CpL = 4.217 kJ/kg K

uL = 279x10⁻⁶Ns/m²

KL = 0.68 W/m K

σ = 58.9x10³N/m

When the water boils on the surface its heat flux is:

$q=0.149h_{fg} \rho _{v} (\frac{\sigma (\rho _{L}-\rho _{v})}{\rho _{v}^{2} } )^{1/4} =0.149*2257*0.596*(\frac{58.9x10^{-3}*(957.9-0.596) }{0.596^{2} } )^{1/4} =18703.42W/m^{2}$

For copper-water, the properties are:

Cfg = 0.0128

The heat flux is:

qn = 0.9 * 18703.42 = 16833.078 W/m²

$q_{n} =uK(\frac{g(\rho_{L}-\rho _{v}) }{\sigma })^{1/2} (\frac{c_{pL}*deltaT }{c_{fg}h_{fg}Pr } \\16833.078=279x10^{-6} *2257x10^{3} (\frac{9.8*(957.9-0.596)}{0.596} )^{1/2} *(\frac{4.127x10^{3}*delta-T }{0.0128*2257x10^{3}*1.76 } )^{3} \\delta-T=20.4$

The tube surface temperature immediately after installation is:

Tinst = 100 + 20.4 = 120.4°C

For rough surfaces, Cfg = 0.0068. Using the same equation:

ΔT = 10.8°C

The tube surface temperature after prolonged service is:

Tprolo = 100 + 10.8 = 110.8°C

8 0

3 years ago

Just some random stufff

NeTakaya

Answer:

Nice!

Explanation:

5 0

3 years ago