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gladu [14]
3 years ago
6

. In order to prevent injury from inflating air bags, it is recommended that vehicle occupants sit at least __________ inches aw

ay from the air bag.
Engineering
2 answers:
scZoUnD [109]3 years ago
8 0
10 inches is the answer
Umnica [9.8K]3 years ago
8 0
10 inches is the answer to that question sir
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In poor weather, you should __ your following distance
Ratling [72]

Answer:

I think reduce your following distance

5 0
4 years ago
Read 2 more answers
(a) A 10.0-mm-diameter Brinell hardness indenter produced an indentation 2.3 mm in diameter in a steel alloy when a load of 1000
vampirchik [111]

Answer:

(a)  We are asked to compute the Brinell hardness for the given indentation. for HB, where P= 1000 kg, d= 2.3 mm, and D= 10 mm.  

Thus, the Brinell hardness is computed as

HB=2P/\pi D{D-\sqrt{D^2-d^2}

=2*1000hg/\pi (10mm)[10mm-\sqrt{(1000^2-(2.3mm)^2} ]

(b)    This  part  of  the  problem  calls  for  us  to  determine  the  indentation diameter d which  will  yield  a  270  HB  when P=  500  kg.  

d=\sqrt{D^2-[D-\frac{2P}{(HB)\pi D} } ]^2\\=\sqrt{(10mm)^2-[10mm-\frac{2*500}{450( \pi10mm)} } ]^2

6 0
4 years ago
Consider a 400 mm × 400 mm window in an aircraft. For a temperature difference of 90°C from the inner to the outer surface of th
True [87]
Um I think the answer for this is 10l
5 0
3 years ago
An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar
Dennis_Churaev [7]

Answer:

a) the heat exchanger area required for the evaporator is 11178.236 m²

b) the required flow rate is 1993630.38 kg/s

Explanation:

Given the data in the question;

Water temperature near the surface = 300 K

temperature at reasonable depths ( cold ) = 280 K

power plant output W' = 2 MW

efficiency η = 3% = 0.03

we know that; efficiency η = W'_{power-out / Q_{supplied

we substitute

0.03 = 2 / Q_{supplied

Q_{supplied = 2 / 0.03

Q_{supplied = 66.667 MW = 66.667 × 10⁶ Watt

Th_{in = 300 K       Th_{out = 292 K

Tc_{in = 290 K       Tc_{out = 290 K    

Now, Heat transfer in evaporator;

Q = UA( LMTD )

so

LMTD = (ΔT₁ - ΔT₂) / ln( ΔT₁ / ΔT₂ )

first we get ΔT₁ and ΔT₂

ΔT₁ = Th_{in - Tc_{out  = 300 - 290 = 10 K

ΔT₂ = Th_{out - Tc_{in  = 292 - 290 = 2 K

so we substitute into our equation;

LMTD = (10 - 2) / ln( 10 / 2 )

LMTD = 8 / ln( 5 )

LMTD = 8 / 1.6094379

LMTD = 4.97

a) Heat transfer Area will be;

Q_H = UA( LMTD )

we substitute

66.667 × 10⁶ = 1200 × A × 4.97

66.667 × 10⁶  = 5964 × A

A = (66.667 × 10⁶) / 5964

A = 11178.236 m²

Therefore, the heat exchanger area required for the evaporator is 11178.236 m²

b) Flow rate  

we know that;

Q_H = m'C_P( T_{in - T_{out )  

specific heat capacity of water Cp = 4.18 (kJ/kg∙°C)

we substitute

66.667 × 10⁶ = m' × 4.18 × ( 300 - 292 )

66.667 × 10⁶ = m' × 33.44

m' = ( 66.667 × 10⁶ ) / 33.44

m' = 1993630.38 kg/s

Therefore, the required flow rate is 1993630.38 kg/s

7 0
3 years ago
Problem Statement: Air flows at a rate of 0.1 kg/s through a device as shown below. The pressure and temperature of the air at l
Tema [17]

Answer:

The answer is "+9.05 kw"

Explanation:

In the given question some information is missing which can be given in the following attachment.

The solution to this question can be defined as follows:

let assume that flow is from 1 to 2 then

Q= 1kw

m=0.1 kg/s

From the steady flow energy equation is:

m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\

If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.

8 0
3 years ago
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