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stealth61 [152]
2 years ago
8

What's the best way to find the load capacity of a crane? Select the best option. Call the manufacturer Ask co-workers Look at t

he load capacity chart in the cab It's best determined by lifting a load
Engineering
1 answer:
Orlov [11]2 years ago
5 0

Answer: You need to lift a load of 15 tons (30,000 pounds) a distance of 25 feet. The distance is measured from the center pin of the crane to the center of the load. Once you determine the distance, look on that line for the largest capacity; that will indicate how many feet of boom must be extended

Explanation:

You might be interested in
A thick steel slab ( 7800 kg/m3 , c 480 J/kg K, k 50 W/m K) is initially at 300 C and is cooled by water jets impinging on one o
Nutka1998 [239]

Answer:

1791 secs  ≈ 29.85 minutes

Explanation:

( Initial temperature of slab )  T1 = 300° C

temperature of water ( Ts ) = 25°C

T2 ( final temp of slab ) = 50°C

distance between slab and water jet = 25 mm

<u>Determine how long it will take to reach T2</u>

First calculate the thermal diffusivity

∝  = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s

<u>next express Temp as a function of time </u>

T( 25 mm , t ) = 50°C

next calculate the time required for the slab to reach 50°C at a distance of 25mm

attached below is the remaining part of the detailed solution

5 0
3 years ago
Drum brakes are usually designed so that the condition of the lining can be checked even if the drum has not been
artcher [175]

Answer:

no it has to be removed

Explanation:

8 0
3 years ago
g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0
Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵  * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α  * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0
2 years ago
What is the difference between a refrigeration cycle and a heat pump cycle?
sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0
3 years ago
Calculate the reluctance of a 4-meter long toroidal coil made of low-carbon steel with an inner radius of 1.75 cm and an outer r
My name is Ann [436]

Answer:

R = 31.9 x 10^(6) At/Wb

So option A is correct

Explanation:

Reluctance is obtained by dividing the length of the magnetic path L by the permeability times the cross-sectional area A

Thus; R = L/μA,

Now from the question,

L = 4m

r_1 = 1.75cm = 0.0175m

r_2 = 2.2cm = 0.022m

So Area will be A_2 - A_1

Thus = π(r_2)² - π(r_1)²

A = π(0.0225)² - π(0.0175)²

A = π[0.0002]

A = 6.28 x 10^(-4) m²

We are given that;

L = 4m

μ_steel = 2 x 10^(-4) Wb/At - m

Thus, reluctance is calculated as;

R = 4/(2 x 10^(-4) x 6.28x 10^(-4))

R = 0.319 x 10^(8) At/Wb

R = 31.9 x 10^(6) At/Wb

8 0
3 years ago
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