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swat32
3 years ago
10

True or False: The differential lock in an AWD-equipped vehicle can be used at any time.

Engineering
2 answers:
Bingel [31]3 years ago
7 0

the answer would be false

Tamiku [17]3 years ago
6 0

Answer:true

Explanation:becuase yes

You might be interested in
Chemical milling is used in an aircraft plant to create pockets in wing sections made of an aluminum alloy. The starting thickne
Lelu [443]

Answer:

a) metal removal rate is 1915.37 mm³/min

b) the time required to etch to the specified depth is 500 min or 8.333 hrs

Explanation:

Given the data in the question;

starting thickness of one work part of interest = 20 mm

depth of series of rectangular-shaped pockets = 12 mm

dimension of pocket = 200 mm by 400 mm

radius of corners of each rectangle = 15 mm

penetration rate = 0.024 mm/minute

etch factor = 1.75

a)

To get the metal removal rate MRR;

The initial area will be smaller compare to the given dimensions of 200mm by 400mm and the metal removal rate would increase during the cut as area is increased. so'

A = 200 × 400 - ( 30 × 30 - ( π × 15² ) )

= 80000 - ( 900 - 707 )      

= 80000 - 193

A = 79807 mm²

Hence, metal removal rate MRR = penetration rate × A

MRR = 0.024 mm/minute × 79807 mm²

MRR = 1915.37 mm³/min

Therefore, metal removal rate is 1915.37 mm³/min

b) To get the time required to etch to the specified depth;

Time to machine ( etch ) =  depth of series of rectangular-shaped pockets / penetration rate

we substitute

Time to machine ( etch ) = 12 mm / 0.024 mm/minute

Time to machine ( etch ) = 500 min or 8.333 hrs

Therefore, the time required to etch to the specified depth is 500 min or 8.333 hrs

3 0
3 years ago
In order to break even, your minimum selling price must be __________ your variable costs.
Gelneren [198K]
Your minimum selling price must be HIGHER THAN your variable costs
7 0
3 years ago
An inventor claims to have invented a heat engine that operates between the temperatures of 627°C and 27°C with a thermal effici
Oksana_A [137]

Answer Explanation:

the efficiency of the the engine is given by=1-\frac{T_2}{T_1}

where T₂= lower temperature

           T₁= Higher temperature

we have given efficiency =70%

lower temperature T₂=27°C=273+27=300K

higher temperature T₁=627°C=273+627=900K

efficiency=1-\frac{T_2}{T_1}

                =1-\frac{300}{900}

                 =1-0.3333

                 =0.6666

                 =66%

66% is less than 70% so so inventor claim is wrong

3 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel.The water depths on either side of the jump are 1 ft and 10 ft.Please calcu
krek1111 [17]

Answer:

a) 42.08 ft/sec

b) 3366.33 ft³/sec

c) 0.235

d) 18.225 ft

e) 3.80 ft

Explanation:

Given:

b = 80ft

y1 = 1 ft

y2 = 10ft

a) Let's take the formula:

\frac{y2}{y1} = \frac{1}{5} * \sqrt{1 + 8f^2 - 1}

10*2 = \sqrt{1 + 8f^2 - 1

1 + 8f² = (20+1)²

= 8f² = 440

f² = 55

f = 7.416

For velocity of the faster moving flow, we have :

\frac{V_1}{\sqrt{g*y_1}} = 7.416

V_1 = 7.416 *\sqrt{32.2*1}

V1 = 42.08 ft/sec

b) the flow rate will be calculated as

Q = VA

VA = V1 * b *y1

= 42.08 * 80 * 1

= 3366.66 ft³/sec

c) The Froude number of the sub-critical flow.

V2.A2 = 3366.66

Where A2 = 80ft * 10ft

Solving for V2, we have:

V_2 = \frac{3666.66}{80*10}

= 4.208 ft/sec

Froude number, F2 =

\frac{V_2}{g*y_2} = \frac{4.208}{32.2*10}

F2 = 0.235

d) El = \frac{(y_2 - y_1)^3}{4*y_1*y_2}

El = \frac{(10-1)^3}{4*1*10}

= \frac{9^3}{40}

= 18.225ft

e) for critical depth, we use :

y_c = [\frac{(\frac{3366.66}{80})^2}{32.2}]^1^/^3

= 3.80 ft

7 0
3 years ago
Read 2 more answers
Marks
Lina20 [59]

Answer:

i think C

Explanation:

if not really sorry

3 0
2 years ago
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