Answer:
6.6 kilo volts = 6.6 k volts
Explanation:
A prefix is a word, number or a letter that is added before another word. In physics we have different prefixes for the exponential powers of 10, that are placed before units in place of those powers. Some examples are:
deci (d) ------ 10⁻¹
centi (c) ------ 10⁻²
milli (m) ------ 10⁻³
kilo (k) ------ 10³
mega (M) ----- 10⁶
giga (G) ------ 10⁹
We have:
6600 volts
converting to exponential form:
=> 6.6 x 10³ volts
Thus, we know that the prefix of kilo (k) is used for 10³.
Hence,
=> <u>6.6 kilo volts = 6.6 k volts</u>
Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.
Answer:
Introduction (including Statement of Problem, Purpose of Research, and Significance of Research): The introduction of a proposal begins with a capsule statement and then proceeds to introduce the subject to a stranger. ... This section is the heart of the proposal and is the primary concern of the technical reviewers.
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