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coldgirl [10]
2 years ago
6

A 10-kg block A is released from rest 2 m above the 5-kg plate P, which can slide freely along the smooth vertical guides BC and

DE. Determine the velocity of the block and plate just after impact. The coefficient of restitution between the block and the plate is e = 0.75. Also, find the maximum compression of the spring due to impact.The spring has an unstretched length of 600 mm
The spring with the 5 kg plate are stationary before impact at 450mm. The 600mm unstretch length of the spring is without the 5 kg plate

Engineering
1 answer:
Liono4ka [1.6K]2 years ago
8 0

Answer:

Check the explanation

Explanation:

<u>Given data </u>

Mass of the block A is Ma = 10kg

Mass of the plate is Mp = 5kg

Coefficient of restitution between the block and that of the plate is e=0.75

Unstretched length of the spring is 0.6m

From the conservation of energy, we have four motion of block A from (1) to (2)

(Ta)_1 + (Va)_1 = (Ta)_2 + (Va)_2

Then accordingly, we will have the step by step calculation in the attached image below .

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7 0
2 years ago
A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)
4vir4ik [10]

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

Your Welcome

8 0
3 years ago
2. There are three drawings that architects and designers use to indicate spaces. What are these drawing?
Zarrin [17]

Answer:

Architectural plans.

Explanation:

An architectural plan is called the drawings made by architects, civil engineers or designers of spaces or interiors, in which these professionals capture their building projects, organizing the distribution of the spaces to be used, the elements to be located in them and, fundamentally, to give construction planning a projection into reality. Thus, the plans help professionals to have a better understanding of the expected end result of the projects they are carrying out.

3 0
3 years ago
A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
SCORPION-xisa [38]

Answer:

F_{thrust} ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

F_{drag} = \frac{1}{2} × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0
3 years ago
Hello , how are yall:))))
SVEN [57.7K]

Answer:

eh I'm good hbu?????????

6 0
2 years ago
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