All categories

2 years ago

IM JI Suneou uo mm

Engineering

1 answer:

Oksi-84 [34.3K]2 years ago

5 0

Answer: g

Explanation:

You might be interested in

QUESTION:

svet-max [94.6K]

OA bloom is smaller than a bar

6 0

2 years ago

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

What is an isentropic process?

snow_tiger [21]

Answer: Isentropic process is the process in fluids which have a constant entropy.

Explanation: The isentropic process is considered as the ideal thermodynamical process and has both adiabatic as well as reversible processes in internal form.This process supports no transfer of heat and no transformation of matter .The entropy of the provided mass also remains unchanged or consistent.These processes are usually carried out on material on the efficient device.

5 0

3 years ago

A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t

gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

$\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}$

using the energy equation for entry and exit value :

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}$

where

$\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}$

$= (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\ V^{2}_{1}\\\\$

$= \frac{1}{(2f (\frac{l}{D}) )} \ V^{2}_{1}\\$

$\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0 =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\$

$\to 0+0+Z_0 = 0 +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g} \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} ) \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})$

L.H.S = R.H.S

7 0

2 years ago

What causes the charging system warning lamp to go out when the engine starts up?

Flura [38]

B is the answer I believe so

7 0

3 years ago