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:

Let the fundamental units be mass= M, time = t, length = L
Then the equation will be in the form


where k is the constant of proportionality.
Now putting the dimensional formula:
![T = KM^{a}L^{b} [LT^{-} ^{2}]^{c}](https://tex.z-dn.net/?f=T%20%3D%20KM%5E%7Ba%7DL%5E%7Bb%7D%20%20%5BLT%5E%7B-%7D%20%5E%7B2%7D%5D%5E%7Bc%7D)

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 = 
where k = 
Answer:
particles of a solid object packed together
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)
]
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%
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
=
...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
=
= 12 kpsi
and
=
...........2
put here value and we get
=
= 17.34 kpsi
now we apply here goodman line equation here that is
...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
solve it we get
FOS = 1.5432