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2 years ago

Are ocean currents always cold

Engineering

1 answer:

swat322 years ago

8 0

Answer:

The surface ocean currents have a strong effect on Earth's climate. ... However, these areas do not constantly get warmer and warmer, because the ocean currents and winds transport the heat from the lower latitudes near the equator to higher latitudes near the poles.

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Impact strips may be designed into a bumper cover.<br> True<br> False

d1i1m1o1n [39]

Answer:

true I think

Explanation:

true I think

5 0

3 years ago

A plate of an alloy steel has a plane-strain fracture toughness of 50 MPa√m. If it is known that the largest surface crack is 0.

Ivahew [28]

Answer:

option B is correct. Fracture will definitely not occur

Explanation:

The formula for fracture toughness is given by;

K_ic = σY√πa

Where,

σ is the applied stress

Y is the dimensionless parameter

a is the crack length.

Let's make σ the subject

So,

σ = [K_ic/Y√πa]

Plugging in the relevant values;

σ = [50/(1.1√π*(0.5 x 10^(-3))]

σ = 1147 MPa

Thus, the material can withstand a stress of 1147 MPa

So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.

8 0

3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

2 years ago

The intake and exhaust processes are not considered in the p-V diagram of Otto cycle. a) true b) false

vovangra [49]

Answer:

b) false

Explanation:

We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.

It consist four processes

1-2:Reversible adiabatic compression

2-3:Constant volume heat addition

3-4:Reversible adiabatic expansion

3-4:Constant volume heat rejection

Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.

But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.

6 0

3 years ago

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about ____

7nadin3 [17]

Answer:

One

For surface-mounted and pendant-hung luminaires, support rods should be placed so that they extend about _one___

<h3>what is supported mounted?</h3>

A structure that holds up or serves as a foundation for something else. Support is a synonym for mounting.

To learn more about it, refer

to brainly.com/question/25689052

#SPJ4

4 0

1 year ago