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3 years ago

11

Technician A says that ultraviolet light is bright, and very hot. Technician B says that ear protection does not need to be used

when welding. Who is correct?
a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B

1 answer:

-Dominant- [34]3 years ago

3 0

Answer:

a. technician a only

Explanation:

I took that worksheet

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A 0.91 m diameter corrugated metal pipe culvert (n = 0.024) has a length of 90 m and a slope of 0.0067. The entrance has a squar

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Answer:

HW=1.71m

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

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2 years ago

A gear motor can develop 2 hp when it turns at 450rpm. If the motor turns a solid shaft with a diameter of 1 in., determine the

kramer

Answer:

Maximum shear stress is;

τ_max = 1427.12 psi

Explanation:

We are given;

Power = 2 HP = 2 × 746 Watts = 1492 W

Angular speed;ω = 450 rev/min = 450 × 2π/60 rad/s = 47.124 rad/s

Diameter;d = 1 in

We know that; power = shear stress × angular speed

So,

P = τω

τ = P/ω

τ = 1492/47.124

τ = 31.66 N.m

Converting this to lb.in, we have;

τ = 280.2146 lb.in

Maximum shear stress is given by the formula;

τ_max = (τ•d/2)/J

J is polar moment of inertia given by the formula; J = πd⁴/32

So,

τ_max = (τ•d/2)/(πd⁴/32)

This reduces to;

τ_max = (16τ)/(πd³)

Plugging in values;

τ_max = (16 × 280.2146)/((π×1³)

τ_max = 1427.12 psi

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3 years ago

if when you put your shirt in your pants, your shirt is tucked, does that mean when your shirt is over your pants, your pants ar

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Answer:

confusing, but yes

Explanation:

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2 years ago

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2 years ago

. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and si

Over [174]

Answer:

y ≈ 2.5

Explanation:

Given data:

bottom width is 3 m

side slope is 1:2

discharge is 10 m^3/s

slope is 0.004

manning roughness coefficient is 0.015

manning equation is written as

$v =1/n R^{2/3} s^{1/2}$

where R is hydraulic radius

S = bed slope

$Q = Av =A 1/n R^{2/3} s^{1/2}$

$A = 1/2 \times (B+B+4y) \times y =(B+2y) y$

$R =\frac{A}{P}$

P is perimeter $= (B+2\sqrt{5} y)$

$R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}$

$Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}$

solving for y $100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}$

solving for y value by using iteration method ,we get

y ≈ 2.5

5 0

3 years ago