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Anon25 [30]
2 years ago
9

Dimension these please.

Engineering
1 answer:
Svetach [21]2 years ago
3 0

Answer:

86

Explanation:

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A(n) _____ is an apparatus that changes alternating current (AC) to direct current (DC)
amid [387]

Answer:

rectifier

Explanation:

7 0
2 years ago
La iluminación de la superficie de un patio amplio es 1600 lx cuando el ángulo de elevación del sol 53°. Calcular la iluminación
gregori [183]

Answer:

 I = 1205.69 Lx

Explanation:

The irradiation or intensity of the solar radiation on the earth is maximum for the vertical fire, with a value I₀

          I = I₀ sin θ

in this case with the initial data we can calculate the initial irradiance

         I₀ = \frac{I}{sin  \ \theta }

         I₀ = 1600 /sin 53

         I₀ = 2003.42 lx

for when the angle is θ = 37º

         I = 2003.42 sin 37

         I = 1205.69 Lx

6 0
3 years ago
If you deposit today 11,613 in an account earning 8% compound interest, for how long should you invest the money in order to ear
Elanso [62]
Y = a (b)^t/p

y is total money

a is original amount

b is growth / decay factor

t is time

p is the frequency of every growth or decay

15131.76 = 11613 x 1.08^x

15131.76 / 11613 = 1.08^x

1.303… = 1.08^x

log1.303…. = xlog1.08

x = 3.43902165741 years
3 0
2 years ago
To put out a class D metal fire, you must _______ the fire.
gladu [14]

To put out a class D metal fire, you must smother the fire and eliminate the oxygen element in the fire.

<h3>What is a Class D fire?</h3>

A class D fire is a type of fire that cannot be extinguished by water. This is because adding water to it reacts with other elements in the fire intensifying the fire even more.

Smothering in this context involves adding a solution like carbon dioxide (CO2) into the fire, this results in a reduction of oxygen in the atmosphere surrounding the class D fire.

By so doing, smothering the fire eliminates the oxygen element in the fire, thereby extinguishing the fire.

You can learn more about extinguishing fires here https://brainly.in/question/760550

#SPJ1

7 0
2 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

Explanation:

The expression for the maximum shear stress is given:

\tau _{M} =\sqrt{(\frac{\sigma _{x}^{2}-\sigma _{y}^{2}  }{2})^{2}+\tau _{xy}^{2}    }

Where

σx = stress in vertical plane = 20 ksi

σy = stress in horizontal plane = -30 ksi

τM = 32 ksi

Replacing:

32=\sqrt{(\frac{20-(-30)}{2} )^{2} +\tau _{xy}^{2}  }

Solving for τxy:

τxy = ±19.98 ksi

The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

σp1 = 20 ksi

σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

Solving both equations:

σp1 = 27 ksi

σp2 = -37 ksi

The shear stress on the vertical plane is zero

4 0
3 years ago
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