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NeX [460]
3 years ago
10

Home safety and security is an _________ process. (7 Letters) Answer

Engineering
1 answer:
Bas_tet [7]3 years ago
5 0

Electric

Answer

Explain

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The period of a pendulum T is assumed to depend only on the mass m, the length of the pendulum `, the acceleration due to gravit
zzz [600]

Answer:

The expression is shown in the explanation below:

Explanation:

Thinking process:

Let the time period of a simple pendulum be given by the expression:

T = \pi \sqrt{\frac{l}{g} }

Let the fundamental units be mass= M, time = t, length = L

Then the equation will be in the form

T = M^{a}l^{b}g^{c}

T = KM^{a}l^{b}g^{c}

where k is the constant of proportionality.

Now putting the dimensional formula:

T = KM^{a}L^{b}  [LT^{-} ^{2}]^{c}

M^{0}L^{0}T^{1} = KM^{a}L^{b+c}

Equating the powers gives:

a = 0

b + c = 0

2c = 1, c = -1/2

b = 1/2

so;

a = 0 , b = 1/2 , c = -1/2

Therefore:

T = KM^{0}l^{\frac{1}{2} } g^{\frac{1}{2} }

T = 2\pi \sqrt{\frac{l}{g} }

where k = 2\pi

8 0
3 years ago
One of the disadvantages of the test of the hypothesis is that the final decision cannot be said to be completely correct eviden
vfiekz [6]

Answer:

A. True

Explanation:

4 0
3 years ago
Can u say what’s this
tatuchka [14]

Answer:

particles of a solid object packed together

7 0
3 years ago
Read 2 more answers
A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP
Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp)\frac{r-1}{r} ]

   0.569   = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569  = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

                                   = 5.2 kPa  

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

   = 0.493 = 49.3%

4 0
3 years ago
• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the
Rudik [331]

Answer:

minimum factor of safety for fatigue is = 1.5432

Explanation:

given data

AISI 1018 steel cold drawn as table

ultimate strength Sut = 63.800 kpsi

yield strength Syt = 53.700 kpsi

modulus of elasticity E = 29.700 kpsi

we get here

\sigma a = \sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}    ...........1

here kb and kt = 1 combined bending and torsion fatigue factor

put here value and we get

\sigma a =  \sqrt{(12 \times 1)^2+3\times (0\times 1)^2}  

\sigma a = 12 kpsi

and

\sigma m = \sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}     ...........2

put here value and we get

\sigma m = \sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}  

\sigma m = 17.34 kpsi

now we apply here goodman line equation here that is

\frac{\sigma m}{Sut} +  \frac{\sigma a}{Se} = \frac{1}{FOS}     ...................3

here Se = 0.5 × Sut

Se = 0.5 × 63.800 = 31.9 kspi

put value in equation 3 we get

\frac{17.34}{63.800} +  \frac{12}{31.9} = \frac{1}{FOS}  

solve it we get

FOS = 1.5432

6 0
3 years ago
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