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sveticcg [70]
3 years ago
6

Consider the two wood pieces that are connected by a velcro as indicated below. The block is subjected to a tension force P and

has dimensions b 2 in and h = 4 in. The velcro can sustain the maximum normal stress Om-1 psi and the maximum shear stress Tn-7 psi. Use α = 20°. What is the maximum value of the force P that can be applied to the block before the velcro breaks connection?

Engineering
2 answers:
sweet [91]3 years ago
7 0

Answer:

Answer is 68.42.

Refer below for the explanation.

Explanation:

Refer to the pictures please.

Amanda [17]3 years ago
3 0

Answer:

P= 68.42 lb

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.I hope my explanation will help you in understanding this particular question.

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A water-filled manometer is used to measure the pressure in an air-filled tank. One leg of the manometer is open to atmosphere.
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P = 150.335\,kPa (Option B)

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What are the major steps in manufacturing biofuels from biomass?
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5 0
2 years ago
A rigid 14-L vessel initially contains a mixture of liquid water and vapor at 100°C with 12.3 percent quality. The mixture is th
tigry1 [53]

Answer:

Q = 65.388 KJ

Explanation:

To calculate the heat required for the given process Q, we recall the energy balance equation.

Therefore, : Q = Δ U = m (u₂ - u₁) ..................equation (1)

We should note that there are no kinetic or potential energy change so the heat input in the system is converted only to internal energy.

Therefore, we will start the equation with the mass of the water (m) using given the initial percentage quality as x₁ = 0.123 and initial temperature t₁ = 100⁰c , we can them determine the initial specific volume v₁ of the mixture. For the calculation, we will also need the specific volume of liquid vₙ  = 0.001043m³/kg and water vapour (vₐ) = 1.6720m³/kg

Therefore, u₁ = vₙ + x₁ . ( vₐ - vₙ)

                   u₁ = 0.001043m³/kg + 0.123 . ( 1.6720m³/kg - 0.001043m³/kg)

                   u₁ = 0.2066m³/kg

Moving forward, the mass of the vapor can then be calculated using the given volume of tank V = 14 L but before the calculation, we need to convert the volume to from liters to m³.

Therefore, V = 14L . 1m² / 1000L = 0.014 m³

Hence, m = V / u₁

                 0.014m³ / 0.2066 m³/kg

              m = 0. 0677 kg

Also, the initial specific internal energy u₁ can be calculated using the given the initial given quality of x₁ , the specific internal energy of liquid water vₐ = 419.06 kj / kg and the specific internal energy of evaporation vₐₙ = 2087.0 kj/kg.

Therefore, u₁ = vₐ + x₁ . vₐₙ

                   u₁ = 419.06 kj / kg + 0.123  .  2087.0 kj/kg

                    u₁ = 675.76 kj/kg

For the final specific internal energy u₂, we first need to calculate the final quality of the mixture x₂ . The tank is rigid meaning the volume does not change and it is also closed meaning the mass does not change.from this, we can conclude the the specific volume also does not change during the process u₁ = u₂. This allows us to use the given final temperature T₂ = 180⁰c to determine the final quality x₂ of the mixture. for the calculation, we will also need the specific volume of liquid vₙ=0.001091m³/kg and vapor vₐ =  0.39248m³/kg

Hence, x₂ = u₂ - vₙ / uₐ

x₂ = 0.2066 m³/kg - 0.001091m³/kg / 0.39248m³/kg

x₂ = 0.524

Moving forward to calculate the final internal energy u₂, we have :

u₂ = vₙ + x₂ . vₙₐ

u₂ = 631.66 kj/kg + 0.524  . 1927.4 kj/kg

u₂ = 1641.62 kj/kg

We now return to equation (1) to plug in the values generated thus far

Q = m (u₂ - u₁)

0. 0677 kg ( 1641.62 kj/kg - 675.76 kj/kg)

Q = 65.388KJ

7 0
3 years ago
Read 2 more answers
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