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3 years ago

What are the equipment requirements for windshields and side windows?

Engineering

1 answer:

lorasvet [3.4K]3 years ago

8 0

Answer:

1.) Safety Glass

-these safety glasses should not be covered with reflective or non-transparent material, and, free of any stickers not required by law.

-the windshields must have safety glass and the side windows should not be covered with any reflective or mirrored appearance and shall have reflection more than 25 percent light

hope it helps..

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A student lives in an apartment with a floor area of 60 m2 and ceiling height of 1.8 m. The apartment has a fresh (outdoor) air

USPshnik [31]

Answer:

Explanation:

5 0

2 years ago

Water exiting the condenser of a power plant at 45 Centers a cooling tower with a mas flow rate of 15,000 kg/s. A stream of cool

MariettaO [177]

Answer: hello your question is incomplete below is the missing part

question :Determine the temperature of the cooled water exiting the cooling tower

answer : T = 43.477° C

Explanation:

Temp of water at exit = 45°C

mass flow rate of cooling tower = 15,000 kg/s

Temp of makeup water = 20°C

Assuming an atmospheric pressure of = 101.3 kPa

Determine temperature of the cooled water exiting the cooling tower

Water entering cooling tower at 45°C

Given that Latent heat of water at 45°C = 43.13 KJ/mol

Cp(wet air) = 1.005+ 1.884(y1)

where: y1 - Inlet mole ratio = (0.01257) / (1 - 0.01257) = 0.01273

Hence : Cp(wet air) = 29.145 + (0.01273) (33.94) = 29.577 KJ/kmol°C

First step : calculate the value of Q

Q = m*Cp*(ΔT) + W(latent heat)

Q = 321.6968 (29.577) (40-30) + 43.13 (18.26089)

Q = 95935.8547 KJ/s

Given that mass rate of water = 15000 kg/s

Hence the temperature of the cooled water can be calculated using the equation below

Q = m*Cp*∆T

Cp(water) = 4.2 KJ/Kg°C

95935.8547 = (15000)*(4.2)*(45 - T)

( 45 - T ) = 95935.8547/ 63000. ∴ T = 43.477° C

5 0

2 years ago

A cylindrical specimen of a metal alloy 45.8 mm long and 9.72 mm in diameter is stressed in tension. A true stress of 378 MPa ca

Sliva [168]

Answer:

390.242 MPa

Explanation:

Attached is the full solution.

8 0

2 years ago

A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground

7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0

3 years ago

A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi

baherus [9]

Answer:

88.18 W

Explanation:

The weight of the boy is given as 108 lb

Change to kg =108*0.453592= 48.988 kg = 49 kg

The slope is given as 6% , change it to degrees as

6/100 =0.06

tan⁻(0.06)= 3.43°

The boy is travelling at a constant speed up the slope = 7mi/hr

Change 7 mi/h to m/s

7*0.44704 =3.13 m/s

Formula for power P=F*v where

P=power output

F=force

v=velocity

Finding force

F=m*g*sin 3.43°

F=49*9.81*sin 3.43° =28.17

Finding the power out

P=28.17*3.13 =88.18 W

4 0

3 years ago