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4 years ago

Please read and answer each question carefully.

Engineering

2 answers:

rodikova [14]4 years ago

7 0

Answer:

correct option is D) Leather gloves

Explanation:

solution

we know that when we work with electric spark then we need such type of glove that protect you from heat and spark

so we use Leather type gloves

because leather can resist sparks and moderate heat

so Leather gloves we used when we do welding work

and The risk of cut and abrasion will also be minimize by the wearing of leather gloves

so that we can say correct option is is D) Leather gloves

BartSMP [9]4 years ago

3 0

Answer: B. and C. are ur top answers if you have any other questions pls ask i hope i helped u with your question

Explanation:

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The B-pillar may also be called the:

slega [8]

Answer:

if you're talking about the car b-post, the answer is "posts"

Explanation:

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3 years ago

PLEASE HURRY!!!

Naily [24]

Answer:

Explanation:

He should get a job in engineering to see what it's like to work in the field.

3 0

3 years ago

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A piece of aluminum wire is 500 ft long and has a diameter of 0.03 inches. What is the resistance of the piece of wire?

dexar [7]

Answer:

8.85 Ω

Explanation:

Resistance of a wire is:

R = ρL/A

where ρ is resistivity of the material,

L is the length of the wire,

and A is the cross sectional area.

For a round wire, A = πr² = ¼πd².

For aluminum, ρ is 2.65×10⁻⁸ Ωm, or 8.69×10⁻⁸ Ωft.

Given L = 500 ft and d = 0.03 in = 0.0025 ft:

R = (8.69×10⁻⁸ Ωft) (500 ft) / (¼π (0.0025 ft)²)

R = 8.85 Ω

5 0

4 years ago

The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero fort < 0. Fort 2 0 they areV =75 ~75e-10

masya89 [10]

Answer:

maximum value of the power delivered to the circuit =3.75W

energy delivered to the element = 3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750

Explanation:

V =75 - 75e-1000t V

l = 50e -IOOOt mA

power = IV = 50 * 10^-3 e -IOOOt * (75 - 75e-1000t)

=50 * 10^-3 e -IOOOt *75 (1 - e-1000t)

maximum value of the power delivered to the circuit =3.75W

the total energy delivered to the element = $\int\limits^t_0 {3.75(e^{ -IOOOt} - e ^{-2OOOt} )} , dx \\\\$

$3750e^{ -IOOOt} - 7000e ^{-2OOOt} -3750$

5 0

3 years ago

The water in a 25-m-deep reservoir is kept inside by a 140-m-wide wall whose cross section is an equilateral triangle as shown i

koban [17]

Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N

(b) 17.1 m

Explanation: The length of wall under the surface can be given by

$b=25m/sin(60)\\=28.867$

The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.

$F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N$

Noting from the Bernoulli equation that

$Po/\rho g sin60 = 100000/1000 * 9.81* sin(60) = 11.77 m \\ \\$

From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:

$Yp = s+\frac{b}{2} +\frac{b^2}{s+\frac{b}{2}+Po/\rho g sin60}= 0+\frac{28.87}{2} +\frac{28.87^2}{0+\frac{28.87}{2}+100000 /1000 *9.81 sin60} = 17.1 m$

Substituting the values gives us the the distance of the surface to be equal to = 17.1 m

7 0

3 years ago