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2 years ago

- Scrap tire management is primarily regulated at the

Engineering

2 answers:

krek1111 [17]2 years ago

5 0

Answer:D

Explanation:

Scrap tire management is primarily regulated at the state

level.

kompoz [17]2 years ago

3 0

Scrap tire management is primarily regulated at the state level.

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MAXImum [283]

C is your answers!!!!!$3&2)//

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3 years ago

Read 2 more answers

A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen

arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

$z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2} }{4g}$ (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

$w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s$

if you consider the equation 1 for the free surface at the edge is equal to

$z_{s(R)} =h_{0} +\frac{w^{2}R^{2} }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft$

7 0

3 years ago

write down your own definition of Engineering, preferably in 4-5 sentences. Maximum of 150 words for your definition???.

ollegr [7]

Answer:

A charge q1=7.0mc is located at the origin and a second charge q2=-5.0mc is located on the x axis, 0.3m the origin find the electric field at the point p which he's coordinates (0,0.40)m

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3 years ago

Air enters the 1 m² inlet of an aircraft engine at 100 kPa and 20° C with a velocity of 180 m/s. Determine: a) The volumetric fl

Shkiper50 [21]

Answer:

a) 180 m³/s

b) 213.4 kg/s

Explanation:

$A_1$ = 1 m²

$P_1$ = 100 kPa

$V_1$ = 180 m/s

Flow rate

$Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s$

Volumetric flow rate = 180 m³/s

Mass flow rate

$\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s$

Mass flow rate = 213.4 kg/s

3 0

3 years ago

A reversible power cycle R and an irreversible power cycle I operate between the same hot and cold thermal reservoirs. Cycle I h

anygoal [31]

Answer: Attached below is the missing diagram

answer :

A) 1) Wr > WI, 2) Qc' > Qc

B) 1) QH' > QH, 2) Qc' > Qc

Explanation:

л = w / QH = 1 - Qc / QH and QH = w + Qc

A) each cycle receives same amount of energy by heat transfer

( Given that ; Л1 = 1/3 ЛR )

1) develops greater bet work 

WR develops greater work ( i.e. Wr > WI )

2) discharges greater energy by heat transfer

Qc' > Qc

solution attached below

B) If Each cycle develops the same net work

1) Receives greater net energy by heat transfer from hot reservoir

QH' > QH ( solution is attached below )

2) discharges greater energy by heat transfer to the cold reservoir

Qc' > Qc

solution attached below

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2 years ago