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kupik [55]
4 years ago
8

The blue sticker on the rear of a vehicle indicates that it's designed

Engineering
2 answers:
Maksim231197 [3]4 years ago
6 0

Its FALSE  123456789012345678987653213e4r5678

GenaCL600 [577]4 years ago
4 0

Answer:

The blue sticker on the rear of a vehicle indicates that it's designed to use compressed natural gas.

Explanation:

You might be interested in
A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and
vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium   K = 205 W/m k

heat flux = k \frac{dt}{dx}

               = 205 \frac{325 - 275}{0.02}

               = 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux = k \frac{dt}{dx}

               = 16.3 \frac{325 - 275}{0.02}

               = 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux = k \frac{dt}{dx}

               = 0.8 \times \frac{325 - 275}{0.02}

               = 2 KW/m2

6 0
3 years ago
For a metal that has a yield strength of 690 MPa and a plane strain fracture toughness (KIc) of 32 MPa-m1/2, compute the minimum
Digiron [165]

Answer:

the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

Explanation:

Given the data in the question;

yield strength σ_y = 690 Mpa

plane strain fracture toughness K_{Ic = 32 MPa-m^{1/2

minimum component thickness for which the condition of plane strain is valid = ?

Now, for plane strain conditions, the minimum thickness required is expressed as;

t ≥ 2.5( K_{Ic / σ_y )²

so we substitute our values into the formula

t ≥ 2.5( 32  / 690  )²

t ≥ 2.5( 0.0463768 )²

t ≥ 2.5 × 0.0021508

t ≥ 0.005377 m or 5.38 mm

Therefore,  the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm

7 0
3 years ago
Suppose there are 93 packets entering a queue at the same time. Each packet is of size 4 MiB. The link transmission rate is 1.4
Ghella [55]

Answer:

0.19s

Explanation:

Queueing delay is the time a job waits in a queue before it can be executed. it is the difference in time betwen when the packet data reaches it destination and the time when it was executed.

Queueing delay =(N-1) L /2R

where N = no of packet =93

L = size of packet = 4MB

R = bandwidth = 1.4Gbps = 1×10⁹ bps

4 MB = 4194304 Bytes

(93 - 1)4194304 / 2× 10⁹

queueing delay =192937984 ×10⁻⁹

=0.19s

5 0
3 years ago
The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas
Lilit [14]

Answer:

A) 209.12 GPa

B) 105.41 GPa

Explanation:

We are given;

Modulus of elasticity of the metal; E_m = 67 GPa

Modulus of elasticity of the oxide; E_f = 390 GPa

Composition of oxide particles; V_f = 44% = 0.44

A) Formula for upper bound modulus of elasticity is given as;

E = E_m(1 - V_f) + (E_f × V_f)

Plugging in the relevant values gives;

E = (67(1 - 0.44)) + (390 × 0.44)

E = 209.12 GPa

B) Formula for upper bound modulus of elasticity is given as;

E = 1/[(V_f/E_f) + (1 - V_f)/E_m]

Plugging in the relevant values;

E = 1/((0.44/390) + ((1 - 0.44)/67))

E = 105.41 GPa

4 0
3 years ago
If a seat could sit on a chair would a chair be able to sit on a seat?
mars1129 [50]

Answer:

no

Explanation:

a seat goes on a chair and never goes off. also the chairs legs are to wide to sit on a seat.

5 0
3 years ago
Read 2 more answers
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