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anzhelika [568]
2 years ago
9

When a system of force acts on a point and stays in rest the forces are in​

Engineering
1 answer:
Vlad1618 [11]2 years ago
6 0

Answer:

uniform

Explanation:

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Milk has a density of as much as 64.6 lb/ft3. What is the gage pressure at the bottom of the straw 6.1 inches deep in the milk?
gregori [183]

Answer:

Explanation:

1 inch is 0.0833333feet

6.1 inches is 0.5083 feet

Density = mass/volume

64.6 = mass/0.50833

mass = 64.6 x 0.5083 =32.83618lb

3 0
3 years ago
Match the word with the definition:
aksik [14]

1. Renewable Resources  = (Renewable means you can keep making it) =  resources that can be replenished (such as trees)

2. Nonrenewable Resources  =  ( Nonrenewable means it can't be made once it is used up) = resources that are gone once they are used (such as fossil fuels)

3. Producer  =  ( produces something) = person who makes goods or provides services

4. Consumer  = ( uses something) =   person whose wants are satisfied by using goods and services

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7 0
3 years ago
Read 2 more answers
Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held
77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} }

{T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}}  \right )^{\frac{K-1}{k} } = 280.15 \times \left (9  \right )^{\frac{1.333-1}{1.333} } = 485.03\ K

The heat supplied = \dot {m}_f × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = \dot m · c_p(T_3 - T_2)

\dot m = 20 kg/s

The heat supplied = 20*c_p(T_3 - T_2) = 21,350 kJ/s

c_p = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

\dfrac{T_5}{T_4}  = \dfrac{2}{1.333 + 1}

T₅ = 1208.42*(2/2.333) = 1035.94 K

C_j = \sqrt{\gamma \times R \times T_5} = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = \dot m × C_j = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 1.9 × 10
Fiesta28 [93]

Answer:

Recall the formula for the maximum stress, σₐ = 2σ₀ *√ (α/ρₓ)

where

σ₀ = tensile stress = 140 MPa = 1.40x 10⁸Pa

α = crack length = 3.8 × 10–2 mm = 3.8 x 10⁻⁵m

ρₓ = radius of curvature = 1.9 × 10⁻⁴mm = 1.9 × 10⁻⁷m

Substituting these values into the formula, we can calculate the max stress as

 ====== 2 x 1.40x 10⁸ x √(3.8 x 10⁻⁵/1.9 × 10⁻⁷)

σₐ  = 24.4MPa

6 0
3 years ago
Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio
AlexFokin [52]

Answer:

attached below

Explanation:

4 0
3 years ago
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