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Jlenok [28]
4 years ago
13

Question 5. Our results support the idea that if left to freely oscillate, a system will vibrate at a natural frequency that dep

ends on the system itself, not on the initial push or stimulus that we impart. Based on this reasoning, consider a beam made out of iron and an otherwise identical beam made out of aluminum. How should the natural frequency of vibration of these two beams compare

Physics
1 answer:
Gekata [30.6K]4 years ago
3 0

Answer:

(b) In ideal condition we neglect mass of spring but in real springs mass of spring adds another factor to its time period.

since we are adding a factor of mass to the system, and frequency being inversely proportional to squared root of mass, we can come to a general conclusion that it effectively reduces the natural frequency .

Explanation:

kindly check the attachment for explanation.

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8 0
3 years ago
What is the heat source for rock formation​
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Answer:

Explanation:

The heat source for rock formation

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4 years ago
The greater the mass is in an object, the higher resistance to a change in movement the object will have. Please select the best
Fofino [41]
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3 0
3 years ago
Xenon has an enthalpy of vaporization of 12.6 kJ/mol and a vapor pressure of 1.00 atm at –108.0 °C. What is the vapor pressure o
earnstyle [38]

Answer:

P₁ = 0.0562 atm

Explanation:

Using the Clausius-Clapeyon equation

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁)       ------ (eqn 1)

Step 1: From the question given, we state out the parameters given

P₁ = ?                T₁ = -148.0⁰C

P₂ = 1atm          T₂ = -108.0⁰C

ΔHvap = 12.6kJ/mol      R = 8.314J/K.mol

Step 2: Do conversions where necessary for unit consistency since our R value is in J/K.mol

a) convert ⁰C to K

1K = ⁰C + 273.15

T₁ = -148.0⁰C => -148.0⁰C + 273.15

T₁ = 125.15K

T₂ = -108.0⁰C => -108.0⁰C + 273.15

T₂ = 166.15K

b) convert kJ/mol to Joules

ΔHvap = 12.6kJ/mol = 12600Joules

substituting parameters into eqn 1

ln (P₁ / P₂) = ΔHvap/R (1/T₂ - 1/T₁)

ln (P₁/1atm) = 12600J / 8.314 (1/166.15 - 1/125.15)

                  = 1515.51 (0.0060 - 0.0079)

                  = 1515.51(-0.0019)

ln (P₁/1atm) = -2.8794

taking exponential of both sides to get rid of the natural log

P₁ = e^ -2.8794

P₁ = 0.05616 atm

P₁ = 0.0562atm

Key Words

1) Clausius-Clapeyon: shows the relationship between pressure and temperature and it is used to estimate the vapour of a solution at a different temperature

5 0
3 years ago
. A student times a car traveling a distance of 2 m. She finds that it takes the car 5 s to
AveGali [126]
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3 0
3 years ago
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