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Zanzabum
4 years ago
7

Help me! Phone Phoebe on 07375410044.

Engineering
2 answers:
natulia [17]4 years ago
4 0

Answer:

yes maam

Explanation:

Step2247 [10]4 years ago
4 0
Answer is yes

Explanation
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A simple undamped spring-mass system is set into motion from rest by giving it an initial velocity of 100 mm/s. It oscillates wi
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Answer:

f=1.59 Hz

Explanation:

Given that

Simple undamped system means ,system does not consists any damper.If system consists damper then it is damped spring mass system.

Velocity = 100 mm/s

Maximum amplitude = 10 mm

We know that for a simple undamped system spring mass system

V_{max}=\omega A

now by putting the values

V_{max}=\omega A

100 = ω x 10

ω = 10 rad/s

We also know that

ω=2π f

10 = 2 x π x f

f=1.59 Hz

So the natural frequency will be f=1.59 Hz.

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4 years ago
Are currently supporting communities affected by GBV in community violations​
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Need help finding the pats of a transformer (Electronics Science) Any help would be appreciated.
Yakvenalex [24]

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21 coils

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3 years ago
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What is the meaning of the measurement met?
san4es73 [151]

Answer: MET(Metabolic equivalent of task) is the measurement technique that is based on the ratio of the energy that is being spend by an individual by any physical activity to the energy spent by individual sitting in quiet position.

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6 0
3 years ago
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical
Olin [163]

Answer:

The theoretical density for Niobium is 1.87 g/cm^3.

Explanation:

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times a^{3}}

where,

\rho = density  of the unit cell

Z = number of atom in unit cell

M = atomic mass

(N_{A}) = Avogadro's number  

a = edge length of unit cell

We have :

Z = 2 (BCC)

M = 92.91 g/mol ( Niobium)

Atomic radius for niobium = r = 0.143 nm

Edge length of the unit cell = a

r = 0.866 a (BCC unit cell)

a=\frac{0.143 nm}{0.866}=0.165 nm=0.165 \times 10^{-7} cm

1 nm = 10^{-7} cm

On substituting all the given values , we will get the value of 'a'.

\rho=\frac{2\times 92.91}{6.022\times 10^{23} mol^{-1}\times (0.165 \times 10^{-7} cm)^{3}}

\rho =1.87 g/cm^3

The theoretical density for Niobium is 1.87 g/cm^3.

6 0
3 years ago
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