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3 years ago

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)

Mass Flow Rate = 0.27 kg/s = 270 g/s

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

V2 = 76.1 m/s

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

P2 = 694962.25 Pa = 695 KPa

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