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Phantasy [73]
4 years ago
7

Given a force of 72 lbs at a distance of 15 ft, calculate the moment produced.​

Engineering
1 answer:
Elis [28]4 years ago
5 0

Answer:

1425.78 N.m

Explanation:

Moments of force is calculated as ;

Moments= Force * distance

M= F*d

The S.I unit for moment of force is Newton-meter (N.m)

Given in the question;

Force = 72 lbs

1 pound = 4.45 N

72 lbs = 4.45 * 72=320.4 N

Distance= 15 ft

1ft= 0.3048 m

15 ft = 15*0.3048 = 4.57 m

d= 4.57 m

M= F*d

M=320.4*4.57 =1425.78 N.m

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Which of the following is a safety device meant to protect an operator from the point of operation?
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Answer:

machine guarding

Explanation:

Machine guarding protects an operator from the point of operation.

6 0
4 years ago
The real power delivered by a source to two impedances, ????1=4+????5⁡Ω and ????2=10⁡Ω connected in parallel, is 1000 W. Determi
kirza4 [7]

Answer:

The question is incomplete, below is the complete question

"The real power delivered by a source to two impedance, Z1=4+j5⁡Ω and Z2=10⁡Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."

answer:

a. 615W, 384.4W

b. 17.4A

Explanation:

To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.

recall that the symbol for admittance is Y and express as

Y=\frac{1}{Z}

Hence for each we have,  

Y_{1} =1/Zx_{1}\\Y_{1} =\frac{1}{4+j5}\\converting to polar \\  Y_{1} =\frac{1}{6.4\leq 51.3}\\  Y_{1} =(0.16 \leq -51.3)S

for the second impedance we have

Y_{2}=\frac{1}{10}\\Y_{2}=0.1S

we also determine the voltage cross the impedance,

P=V^2(Y1 +Y2)

V=\sqrt{\frac{P}{Y_{1}+Y_{2}}}\\

V=\sqrt{\frac{1000}{0.16+0.1}}\\ V=62v

The real power in the impedance is calculated as

P_{1}=v^{2}G_{1}\\P_{1}=62*62*0.16\\ P_{1}=615W

for the second impedance

P_{2}=v^{2}*G_{2}\\   P_{2}=62*62*0.1\\384.4w

b. We determine the equivalent admittance

Y_{total}=Y_{1}+Y_{2}\\Y_{total}=(0.16\leq -51.3 )+0.1\\Y_{total}=(0.16-j1.0)+0.1\\Y_{total}=0.26-J1.0\\

We convert the equivalent admittance back into the polar form

Y_{total}=0.28\leq -19.65\\

the source current flows is

I_{s}=VY_{total}\\I_{s}=62*0.28\\I_{s}=17.4A

6 0
3 years ago
Starting the vehicle's engine and listening to its operation can
miskamm [114]
I pretty sure it’s true because you can start your car and it sounds different that probably means something is wrong
3 0
3 years ago
As temperature decreases a batteries availability power what
SSSSS [86.1K]
The answer is increases because when something like that decreases it’s always decreasing that probly makes no sense Imao but it’s increases
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3 years ago
The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f
kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

C_{v} =AT^{3}

Substituting in the expression of the question:

C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3}  } )T^{3}

\theta _{D} =(\frac{12\pi ^{4}RT^{3}  }{5C_{v}   } ) ^{1/3}

Here

C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K

Replacing:

\theta _{D}  =(\frac{12\pi ^{4}*8.31*15^{3}  }{5*0.1241} )^{1/3} =375.2361K

3 0
3 years ago
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