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4 years ago

11

You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac

tivity B. it’s harder to see people C. insurance is more expensive D. your car uses more gas

1 answer:

torisob [31]4 years ago

5 0

Answer:

B, its the only valid answer

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Who is ACP pradyuman

Kamila [148]

Answer:

he is the chief officer of crime invistegating department.

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3 years ago

Read 2 more answers

Discuss with your neighbor your brain as a computer.

Misha Larkins [42]

Senors are a type of device that produce a amount of change to the output to a known input stimulus.
Input signals are signals that receive data by the system and outputs the ones who are sent from it. Hope this helps ;)

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3 years ago

What should a technician do before entering a confined space? Question 1 options: A) Post another worker outside the confined sp

Lady bird [3.3K]

Answer:

C) Wear clothes that are resistant to spontaneous arcing.

Explanation:

A confined area or space is one that has strict restriction against movement of people due to the availability of hazardous substance or material within the confined space. If there is need for anyone to enter the space, it is very important that the person make use of personal protective equipment (PPE).

Personal protective equipment are materials that are capable of inhibiting injury or hazard when used appropriately. Examples are: helmet, respirators, gloves, goggles, ear muffs etc.

Before a technician should enter a confined space, he/ she should use a PPE by wearing clothes that are resistant to spontaneous arcing.

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3 years ago

Technician A says that after replacing a power steering hose, the system should be flushed, refilled, and bled. Technician B say

dedylja [7]

The. Answer will be D

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4 years ago

The pressure distribution over a section of a two-dimensional wing at 4 degrees of incidence may be approximated as follows: Upp

Aliun [14]

Answer:

The lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

Explanation:

The Upper Surface Cp is given as

$Cp_u=-0.8 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.8*0.6+0.4*0.1$

The Lower Surface Cp is given as

$Cp_l=-0.4 *0.6 +0.1 \int\limits^1_{0.6} \, dx =-0.4*0.6+0.4*0.1$

The difference of the Cp over the airfoil is given as

$\Delta Cp=Cp_l-Cp_u\\\Delta Cp=-0.4*0.6+0.4*0.1-(-0.8*0.6-0.4*0.1)\\\Delta Cp=-0.4*0.6+0.4*0.1+0.8*0.6+0.4*0.1\\\Delta Cp=0.4*0.6+0.4*0.2\\\Delta Cp=0.32$

Now the Lift Coefficient is given as

$C_L=\Delta C_p cos(\alpha_i)\\C_L=0.32\times cos(4*\frac{\pi}{180})\\C_L=0.3192$

Now the coefficient of moment about the leading edge is given as

$C_M=-0.3*0.4*0.6-(0.6+\dfrac{0.4}{3})*0.2*0.4\\C_M=-0.1306$

So the lift coefficient is 0.3192 while that of the moment about the leading edge is-0.1306.

5 0

3 years ago