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VashaNatasha [74]
3 years ago
14

A forklift raises a 90.5 kg crate 1.80 m. (a) Showing all your work and using unity conversion ratios, calculate the work done b

y the forklift on the crane, in units of kJ. (b) If it takes 12.3 seconds to lift the crate, calculate the useful power supplied to the crate in kilowatts.
Engineering
2 answers:
klasskru [66]3 years ago
5 0

Answer:

(a) Work done is 1.59642 kJ

(b) Useful power supplied = 0.1298 kW

Explanation:

(a) Work done = mass of crate × acceleration due to gravity × distance = 90.5 kg × 9.8 m/s^2 × 1.8 m = 1596.42 kgm^2/s^2 × 1 J/1 kgm^2/s^2 × 1 kJ/1000 J = 1.59642 kJ

(b) Useful power supplied = work done ÷ time = 1.59642 kJ ÷ 12.3 s = 0.1298 kJ/s × 1 kW/1 kJ/s = 0.1298 kW

KengaRu [80]3 years ago
5 0
<h2>Answer:</h2>

(a) 1.629 kJ

(b) 0.132 kW

<h2>Explanation:</h2>

The workdone (W) on a body is the product of the force (F) acting on the body and the displacement (s) caused by the force. This can be written as;

W = F x s             ----------------(i)

Where;

F = m x a     [From Newton's law]          ------------------(ii)

m = mass of the body

a = acceleration of the body

Since the direction of motion is upwards, the acceleration acting on the body is the one due to gravity.

i.e a = g = 10m/s²

From the question, the mass of the crate is 90.5kg.

Substitute these values into equation (ii) to get the force acting on the body as follows;

F = 90.5 x 10

F = 905N

From the question;

The displacement (s) of the crate is 1.80m

Substitute the values of the force F and the displacement s into equation (i) as follows;

W = 905 x 1.8

W = 1629 J

Convert the result to kJ by dividing by 1000 as follows;

W = (1629/1000) kJ

W = 1.629 kJ

Therefore, the work done on the crate is 1.629 kJ

(b) Power (P) is the ratio of work done (W) to the time taken (t) to do the work. i.e;

P = W / t            -----------------(ii)

From the question;

t = 12.3s

Also;

W = 1.629 kJ   [as calculated in a]

Substitute these values into equation(ii) as follows;

P = 1.629 / 12.3

P = 0.132 kW

Therefore the useful power supplied to the crate is 0.132 kW

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Sergio039 [100]

Answer:

The ten numbers to be filled in the blanks are: 18, 7, 7, 11, 18, 36, 3, 8, 13, 50.

Explanation:

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now at Node 18;

left = right = 0; hence condition is not satisfied

18 is printed first.

the value 18 is returned .

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from there we move to 7;

just like 18, similar things happen with 7 and 7 is printed, the value 7 is returned.

Now coming to Node 4,

left = 0, right = 7 ; hence the condition is satisfied & res = 7; 7 is printed.

For Node 16, left = 7 ; right = 11(but for this we visit 11 first and 11 is printed)

for 16; condition is satisfied; res = 7 + 11 = 18 ; 18 is printed

Now for 5; left = right = 18; the condition is satisfied; so res = 18 + 18 = 36; 36 is printed

Next we visit Node 3; 3 is printed & 3 is returned

Then Node 8 ; 8 is printed & 8 is returned

for Node 13; left = 3, right = 8 ; condition is not satisfied, 13 is printed.

For Node 50; left = 36 right = 13 ; condition is not satisfied hence 50 is printed.

So the order of printing is  18 7 7 11 18 36 3 8 13 50.

4 0
4 years ago
If the head loss in a 30 m of length of a 75-mm-diameter pipe is 7.6 m for a given flow rate of water, what is the total drag fo
Stolb23 [73]

Answer:

526.5 KN

Explanation:

The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.

But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.

h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg

where ρ = density of the fluid and g = acceleration due to gravity

h = ΔP/ρg

ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa

Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with

Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa

Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²

Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN

3 0
4 years ago
How many watts are consumed in a circuit having a power factor of 0. 2 if the input is 100 vac at 4 amperes?.
Anna71 [15]

The watts that are consumed is 80 watts.

<h3>What power factor?</h3>

The term power factor has to do with the measure of the efficiency of the use of energy. Recall that power is defined as the rate of doing work. The magnitude of the power factor shows the extent to which the power is used.

Now, to obtain the watts are consumed in a circuit having a power factor of 0. 2 if the input is 100 vac at 4 amperes we have;  V × I × PF = 100V × 4A × 0.2 = 80 watts.

Learn more about power factor:brainly.com/question/10634193

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6 0
2 years ago
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some measurements indicate that the depth of the unknown
barxatty [35]

Answer:

The specific weight of unknown liquid is found to be 15 KN/m³

Explanation:

The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.

Total Pressure = Pressure of oil + Pressure of unknown liquid

65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)

65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)

(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²

(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m

<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>  

4 0
3 years ago
For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _______ and using
serious [3.7K]

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

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