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emmainna [20.7K]
3 years ago
7

Why do you think the Combination Square is better to use than the tape

Engineering
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

The correct answer to the following question will be "Because the head acts as a stop".

Explanation:

  • Since the head serves as either a stopping (especially in comparison to something like a floppy tape clip or hanger), the combination square seems to be more reliable and accurate than those of the tape measure.
  • And maybe you'll know that certain style outlines are still square to something like the sides or finishes.

So that the above seems to be the right answer.

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Which is not an affect of alcohol
Gre4nikov [31]
Pooping problems is not an affect
6 0
3 years ago
In a planetary geartrain with a form factor of 8, the sun gear rotates clockwise at 5 rad⁄s and the ring gear rotates clockwise
lina2011 [118]

Answer:

D. N= 11. 22 rad/s (CW)

Explanation:

Given that

Form factor R = 8

Speed of sun gear = 5 rad/s (CW)

Speed of ring gear = 12 rad/s (CW)

Lets take speed of carrier gear is N

From Algebraic method ,the relationship between speed and form factor given as follows

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

here negative sign means that ring and sun gear rotates in opposite direction

Lets take CW as positive and ACW as negative.

Now by putting the values

\dfrac{N_{sun}-N}{N_{ring}-N}=-R

\dfrac{5-N}{12-N}=-8

N= 11. 22 rad/s (CW)

So the speed of carrier gear is 11.22 rad/s clockwise.

8 0
3 years ago
Air flows through a convergent-divergent duct with an inlet area of 5 cm² and an exit area of 3.8 cm². At the inlet section, the
Luda [366]

Answer:

The mass flow rate is 0.27 kg/s

The exit velocity is 76.1 m/s

The exit pressure is 695 KPa

Explanation:

Assuming the flow to be steady state and the behavior of air as an ideal gas.

The mass flow rate of the air is given as:

Mass Flow Rate = ρ x A1 x V1

where,

ρ = density of air

A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²

V1 = inlet velocity = 100 m/s

For density using general gas equation:

PV = nRT

PV = (m/M)RT

PM/RT = ρ

ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k

ρ = 7.11 kg/m³

Therefore,

Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)

<u>Mass Flow Rate = 0.27 kg/s = 270 g/s</u>

Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:

Mass Flow Rate = ρ x A2 x V2

where,

ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)

A2 = exit area = 5 cm² = 5 x 10^-4 m²

V2 = exit velocity = ?

Therefore:

0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2

<u>V2 = 76.1 m/s</u>

Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:

P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2

Since, both inlet and exit are at same temperature.

Therefore, h1 = h2, and those terms will cancel out.

P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²

P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²

P2 = P1 + (1/2) ρ (V1² - V2²)

P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]

P2 = 680000 Pa + 14962.25 Pa

<u>P2 = 694962.25 Pa = 695 KPa</u>

4 0
3 years ago
A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted
alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - D^{2}/Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf =  free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

    = 240v/h

q = Dj ( U - U^{2}/Uf)

2100 = 240 ( U - U^{2}/55)

Solve for U

U = 44m/h

5 0
3 years ago
How can you organize a PowerPoint presentation so it engages the audience?
Yakvenalex [24]

Answer:

Use a story structure,

Explanation: Usually story engage the audience.

7 0
3 years ago
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