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Sindrei [870]
3 years ago
15

Need a little help here :(

Physics
2 answers:
Goshia [24]3 years ago
8 0

Answer:

The output out be 200

Explanation:

Hope this helps :))

CaHeK987 [17]3 years ago
5 0

Answer:

1600 lb

Explanation:

The pressure on Ram Input A is the same pressure on the Output Ram.

P₁ = P₂

F₁ / A₁ = F₂ / A₂

F₁ / (π d₁²/4) = F₂ / (π d₂² / 4)

F₁ / d₁² = F₂ / d₂²

(100 lb) / (0.5 in)² = F / (2 in)²

F = 1600 lb

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Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, \rho = 10.49 g/cm^{3}

Density of Pd, \rho = 12.02 g/cm^{3}

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, \rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}

where

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\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}

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Average atomic weight is given as:

A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}

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A_{Pd} = 106

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A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78

In the similar way, average density is given as:

\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}

\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}

Therefore, edge length is given by eqn (1) as:

a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm

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