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3 years ago

Identify five safety hazards that should be included in the design of the school

Engineering

1 answer:

erik [133]3 years ago

7 0

Answer:

1) function of fire doors and making sure theyre properly wired to fire alarms

2) proper water piping and purifaction for water fountains and sinks.

3) falty sprinkler systems/rsuty sprikler systems that wont work

4) weather durable roofing and walls for storms and snow depending on were your located .

5) rsuted pipes in showers or fountains that could give you tetnus or other disaeases

Explanation:

HOPE THIS HELPS good luck!!

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Take water density and kinematic viscosity as p=1000 kg/m3 and v= 1x10^-6 m^2/s. (c) Water flows through an orifice plate with a

guapka [62]

Answer:

$K_v=12.34$

Explanation:

Given;

For orifice, loss coefficient, K₀ = 10

Diameter, D₀ = 45 mm = 0.045 m

loss coefficient of the orifice, Ko = 10

Diameter of the gate valve, Dy = 1.5D₀ = 1.5 × 0.045 m = 0.0675 m

Total head drop, Δhtotal=25 m

Discharge, Q = 10 l/s = 0.01 m³/s

Now,

the velocity of flow through orifice, Vo = Discharge / area of the orifice

Vo = $\frac{0.01}{\frac{\pi}{4}0.045^2}$

Vo = 6.28 m/s

also,

the velocity of flow through gate valve, $V_v$ = Discharge / area of the orifice

$V_v$ = $\frac{0.01}{\frac{\pi}{4}0.0675^2}$

$V_v$ = 2.79 m/s

Now,

the total head drop = head drop at orifice + head drop at gate valve

25 m = $K_o\frac{V_o^2}{2g}+K_v\frac{V_v^2}{2g}$

where,

$K_v$ is the loss coefficient for the gate valve

on substituting the values, we get

25 m = $10\frac{6.28^2}{2\times 9.81}+K_v\frac{2.79^2}{2\times9.81}$

$K_v\frac{2.79^2}{2\times9.81}$ = 4.898

$K_v=12.34$

3 0

4 years ago

MANUFACTURING QUESTION PLEASE HELP

Nat2105 [25]

Answer:

"He then pours the metal into a mold and allows it to solidify."

3 0

3 years ago

Relation between Poisson's ration, young's modulus, shear modulus for an isotropic material

poizon [28]

Answer:

E=2 G(1+2μ)

Explanation:

Isotropic material :

Those material have same property in all direction is known as isotropic material.

Homogeneous material :

Those material have same property through out the volume is known as homogeneous material.

Relationship between Poisson ratio ,young modulus and shear modulus:

E=2 G(1+2μ)

Relationship between Poisson ratio ,bulk modulus and shear modulus:

E=2 K(1-μ)

Where

E is young modulus.

G is shear modulus.

μ is Poisson ratio.

8 0

3 years ago

Sea water with a density of 1025 kg/m3 flows steadily through a pump at 0.21 m3 /s. The pump inlet is 0.25 m in diameter. At the

myrzilka [38]

Answer:

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

Explanation:

The pump is modelled after applying Principle of Energy Conservation, whose form is:

$\frac{P_{1}}{\rho\cdot g}+ \frac{v_{1}^{2}}{2\cdot g} +z_{1} + h_{pump}=\frac{P_{2}}{\rho\cdot g}+ \frac{v_{2}^{2}}{2\cdot g} +z_{2}$

The head associated with the pump is cleared:

$h_{pump} = \frac{P_{2}-P_{1}}{\rho\cdot g}+\frac{v_{2}^{2}-v_{1}^{2}}{2\cdot g}+(z_{2}-z_{1})$

Inlet and outlet velocities are found:

$v_{1} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.25\,m)^{2} }$

$v_{1} \approx 4.278\,\frac{m}{s}$

$v_{2} = \frac{0.21\,\frac{m^{3}}{s} }{\frac{\pi}{4}\cdot (0.152\,m)^{2} }$

$v_{2} \approx 11.573\,\frac{m}{s}$

Now, the head associated with the pump is finally computed:

$h_{pump} = \frac{175\,kPa-81.326\,kPa}{(1025\,\frac{kg}{m^{3}} )\cdot (9.807\,\frac{m}{s^{2}} )} +\frac{(11.573\,\frac{m}{s} )^{2}-(4.278\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )} + 1.8\,m$

$h_{pump} = 7.705\,m$

The power that pump adds to the fluid is:

$\dot W_{pump} = \dot V \cdot \rho \cdot g \cdot h_{pump}$

$\dot W_{pump} = (0.21\,m^{3})\cdot (1025\,\frac{kg}{m^{3}})\cdot (9.807\,\frac{m}{s^{2}})\cdot(7.705\,m)$

$\dot W_{pump} = 16264.922\,W\,(16.265\,kW)$

4 0

3 years ago

Knowing that v = –8 m/s when t = 0 and v = 8 m/s when t = 2 s, determine the constant k. (Round the final answer to the nearest

docker41 [41]

Answer:

a)We know that acceleration a=dv/dt

So dv/dt=kt^2

dv=kt^2dt

Integrating we get

v(t)=kt^3/3+C

Puttin t=0

-8=C

Putting t=2

8=8k/3-8

k=48/8

k=6

5 0

3 years ago