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anzhelika [568]
3 years ago
5

An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g

ravitational potential energy of the object, in kJ, relative to the surface of the earth.
Engineering
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

Gravitational Potential Energy = mass\times g\times Height

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = 9.78m/s^{2}

Applying values in the equation we get

Gravitational Potential Energy=251X9.78X24 Joules

Gravitational Potential Energy=58914.72 Joules

Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ

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E. None of the above
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2 years ago
If the efficiency of the boiler is 91.2 % , the overall efficiency of the turbine, which includes the Carnot efficiency and its
Tju [1.3M]

Answer:

Net efficiency of generating unit = 42.08 - 5 = 37.08 %

Explanation:

We have given that efficiency of the boiler = 91.2 % = 0.912

Carnot efficiency = 46.9 % = 0.469

Efficiency of generator = 98.4% =0.984

We have to find the efficiency of total generating unit

For finding the efficiency of total generating unit we have to multiply all the efficiencies

So efficiency of generating unit = 0.912×0.469×0.984 = 0.4208 = 42.08 %

For plant losses we have to subtract 5%

So net efficiency of generating unit = 42.08 - 5 = 37.08 %

4 0
3 years ago
A crude fermenter is set up in a shed in the backyard of a suburban house. Under anaerobic conditions with ammonia as the nitrog
Aleks04 [339]

Answer:

using calculations Heat losses will be 4512 J

5 0
3 years ago
A person holds her hand out of an open car window while the car drives through still air at 65 mph. Under standard atmospheric c
Paraphin [41]

Answer:

10.8\ \text{lb/ft^2}

101.96\ \text{lb/ft}^2

Explanation:

v_1 = Velocity of car = 65 mph = 65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}

\rho = Density of air = 0.00237\ \text{slug/ft}^3

v_2=0

P_1=0

h_1=h_2

From Bernoulli's law we have

P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}

The maximum pressure on the girl's hand is 10.8\ \text{lb/ft^2}

Now v_1 = 200 mph = 200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}

P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2

The maximum pressure on the girl's hand is 101.96\ \text{lb/ft}^2

5 0
3 years ago
Consider the following list. list = {24, 20, 10, 75, 70, 18, 60, 35} Suppose that list is sorted using the insertion sort algori
Greeley [361]

Answer:

Option B

10,20,24,75,70,18,60,35

Explanation:

The first, second and third iteration of the loop will be as follows

insertion sort iteration 1: 20,24,10,75,70,18,60,35

insertion sort iteration 2:10,20,24,75,70,18,60,35

insertion sort iteration 3: 10,20,24,75,70,18,60,35

8 0
3 years ago
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