All categories

1 year ago

Which option describes the best material for the project described in the following scenario?

Engineering

2 answers:

il63 [147K]1 year ago

8 0

Answer: Its "D"

Explanation:

Did the same question on EDG 21 for Engineering and Design

padilas [110]1 year ago

7 0

Answer:

I think the answer is:

(4) Plastic is economic, easily cleanable, & moldable

Explanation:

You might be interested in

un contenedor de 0.01m∧3 se llena con 2kg de nitrogeno a una presion de 15mpa ¿cual es la temperatura del nitrogeno?resolver uti

777dan777 [17]

Ecfñnokg pinogdf gabn Etta r

7 0

2 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

In a home, air infiltrates from the outside through cracks around doors and windows. Consider a residence where the total length

masya89 [10]

Answer:

Time period = 41654.08 s

Explanation:

Given data:

Internal volume is 210 m^3

Rate of air infiltration $9.4 \times 10^{-5} kg/s$

length of cracks 62 m

air density = 1.186 kg/m^3

Total rate of air infiltration $= 9.4\times 10^{-5} \times 62 = 582.8\times 10{-5} kg/s$

total volume of air infiltration $= \frac{582.8\times 10{-5}}{1.156} = 5.04\times 10^{-3} m^3/s$

Time period $= \frac{210}{5.04\times 10^{-3}} = 41654.08 s$

3 0

3 years ago

How many electrons move past a fixed reference point every t = 2.55 ps if the current is i = 7.3 μA ? Express your answer as an

iris [78.8K]

Answer:

116.3 electrons

Explanation:

Data provided in the question:

Time, t = 2.55 ps = 2.55 × 10⁻¹² s

Current, i = 7.3 μA = 7.3 × 10⁻⁶ A

Now,

we know,

Charge, Q = it

thus,

Q = (7.3 × 10⁻⁶) × (2.55 × 10⁻¹²)

Q = 18.615 × 10⁻¹⁸ C

Also,

We know

Charge of 1 electron, q = 1.6 × 10⁻¹⁹ C

Therefore,

Number of electrons past a fixed point = Q ÷ q

= [ 18.615 × 10⁻¹⁸ ] ÷ [ 1.6 × 10⁻¹⁹ ]

= 116.3 electrons

4 0

2 years ago

35 points an brainiest if correct

Jet001 [13]

I think the answer is B. 10D

6 0

2 years ago