Ps115L Technical Physics Lab Homework 9 Due At Beginning

Ps115l Technical Physics Labhw9homework 9due At Beginning Of Lab 10

Ps115l Technical Physics Labhw9homework 9due At Beginning Of Lab 10

Ps115l Technical Physics Labhw9homework 9due At Beginning Of Lab 10

PS115L Technical Physics Lab HW#9 Homework 9 Due at beginning of Lab . A mass of .6 kg moving with a velocity of 2.4 m/s in the positive direction experiences a totally inelastic collision with a mass of 1.1 kg. What is the final velocity of the masses? 2. A mass of 7.2 kg moving with a velocity of .8 m/s experiences an elastic collision with a mass that is stationary.

After the collision, the velocity of the second mass is 10.286 m/s while the velocity of the first mass is 0 m/s. What is the mass of the second object? 3. A mass of .43 kg moving with a velocity of 1.5 m/s in the positive direction experiences an elastic collision with another mass of .22 kg moving with a velocity of 3.4 m/s in the negative direction. What will the velocities of the two masses be after the collision?

Remember that in elastic collisions, both kinetic energy and momentum are conserved. The Anderson Corporation (a privately-held company) is interested in in expanding its operating capacity, but is short on cash to finance the expansion. The president of the company does not want to borrow funds, but would not be averse to issuing some of the company’s stock to acquire land and a building. Baker Company owned land and a building, and was interested in disposing of them. After some negotiation, Anderson and Baker agreed to swap the land and building for shares in the Anderson Corporation.

The president of Anderson Corporation has asked you how to account for this transaction, including whether the transaction qualifies as an exception to the general rule to use fair value, and the value to place on the transaction and its components. Required : Research the FASB Accounting Standards Codification and prepare a memo to the president that answers his questions. Provide appropriate citations from the Codification . Hints : While providing the appropriate citations from the Codification, the citations could be either paraphrase or quotation . The standards of FASB Accounting Standards Codification provided in the memo must be cited (For example: ASC ) .

While explaining the way of recording the relevant transactions, please provide corresponding journal entries in the memo. Besides, explain how I determine the amount. PS115L Technical Physics Lab HW#8 Homework 8 Due at beginning of Lab . From the lab, you know that the speed of sound , where is the current temperature. If you measure the wavelength of a 440 Hz sound wave to be .80364 m (the sound wave is moving through air), what is the current temperature?

2. A container is filled halfway up with water, after which it is filled the rest of the way up with oil. Because the liquids have different densities, the water and oil will not mix, so that the oil will sit in a layer on top of the water. You then drop an object with a volume of and a mass of .0768 kg into the container. Explain shortly what will happen to the object (i.e., whether it sinks or floats), why this happens, and show calculations to support your answer. (This need not be an essay or even complete sentences; as long as you include all relevant information in a clear format, it will be good.) 3.

Water flows in a pipe. At first, the pipe has a diameter of .35 m and the water through the pipe has a velocity of 2.2 m/s. Then the pipe widens so that it now has a diameter of .84 m. What is the velocity of the water in the larger pipe? Assume the flow is incompressible and is not viscous.

4. You are using the pressure from a jet of water to maintain the height of a column of water in a tube at 1.8 m. How fast is the water in the jet moving? Use the pressure (energy density) equation used in the lab. 5.

A ship has a mass of and a volume of . How much cargo (in kg) can the ship carry before it sinks?

Paper For Above instruction

The collection of physics problems and accounting scenario provided requires a thorough understanding of collision mechanics, fluid dynamics, acoustics, and financial accounting standards. This paper will analyze each problem systematically and provide detailed solutions, calculations, and accounting explanations based on current academic and professional standards.

Physics Problems and Solutions

1. Inelastic Collision of Two Masses

Given a mass m1 = 0.6 kg moving at v1 = 2.4 m/s and a second mass m2 = 1.1 kg stationary, the final velocity (v_f) after a completely inelastic collision can be found using conservation of momentum:

m1  v1 + m2  v2 = (m1 + m2) * v_f

Since m2 is stationary (v2 = 0), the equation simplifies to:

v_f = (m1  v1) / (m1 + m2) = (0.6  2.4) / (0.6 + 1.1) = 1.44 / 1.7 ≈ 0.847 m/s

Thus, the final velocity of both masses after the inelastic collision is approximately 0.847 m/s in the positive direction.

2. Elastic Collision and Mass of the Second Object

Given m1 = 7.2 kg, initial velocity v1= 0.8 m/s, and the second mass is stationary (v2=0), resulting in post-collision velocities v1' = 0 m/s and v2' = 10.286 m/s.

For elastic collisions, the conservation of momentum and kinetic energy applies:

m1  v1 + m2  v2 = m1  v1' + m2  v2'


(7.2)(0.8) + m2  0 = 7.2  0 + m2 * 10.286

Calculating m2:

5.76 = 10.286 * m2 ⇒ m2 = 5.76 / 10.286 ≈ 0.56 kg

Therefore, the second object has an approximate mass of 0.56 kg.

3. Velocities after elastic collision between two masses

Given m1=0.43 kg, v1=1.5 m/s (positive), m2=0.22 kg, v2=-3.4 m/s. Elastic collisions conserve both momentum and kinetic energy. The velocities after collision (v1', v2') are given by:

v1' = [(m1 - m2)  v1 + 2  m2 * v2] / (m1 + m2)


v2' = [(m2 - m1)  v2 + 2  m1 * v1] / (m1 + m2)

Calculations:

v1' = [(0.43 - 0.22)  1.5 + 2  0.22 * (-3.4)] / (0.43 + 0.22)


= (0.21  1.5 + 0.44  -3.4) / 0.65


= (0.315 - 1.496) / 0.65 ≈ -1.181 / 0.65 ≈ -1.818 m/s


v2' = [(0.22 - 0.43)  (-3.4) + 2  0.43 * 1.5] / 0.65


= (-0.21  -3.4 + 0.86  1.5) / 0.65


= (0.714 + 1.29) / 0.65 ≈ 2.004 / 0.65 ≈ 3.086 m/s

The velocities after the collision are approximately v1' ≈ -1.818 m/s and v2' ≈ 3.086 m/s.

Accounting Scenario: Asset Swap and Financial Reporting

The scenario involves the exchange of land and building for shares in Anderson Corporation. According to the FASB Accounting Standards Codification (ASC 958-605), transactions involving the exchange of assets for stock should generally be measured at fair value unless specific exceptions apply. The key question is whether this transaction qualifies for an exception to the general rule to use fair value.

ASC 958-605-25-17 states that "a nonprofit entity shall measure contributed services at fair value if the services create or enhance nonfinancial assets or are specialized skills that are provided by individuals possessing those skills." Although this is not directly applicable here, the broader principle emphasizes applying fair value measurement for asset exchanges unless there is a specific exception.

In the context of asset exchanges between entities, ASC 845-10-30-1 provides guidance on recognizing gains and losses. Specifically, if an asset is exchanged for stock, and the transaction lacks commercial substance, the fair value approach may be bypassed in certain cases, but generally, fair value measurement is required.

Given the lack of explicit exemption in the ASC for this type of transaction, the fair value approach should be adopted, with the value of the land and building, as well as the corresponding stock issued, recorded at fair value at the date of exchange.

To record the transaction:

  • Debit Land and Building (asset accounts): at fair value
  • Credit Stock (equity account): at fair value
  • Recognize any gain or loss based on the difference between book values and fair values

The valuation process involves appraisals of the land and building to determine their fair market value, which would typically involve third-party valuation reports or recent comparable sales data.

In summary, this transaction should be recorded at fair value, and the standard journal entries involve recognizing the assets acquired at their appraisal values and issuing stock at its fair value based on the stock’s market price at the time of exchange.

Fluid Dynamics and Other Physics Problems

Speed of Sound and Temperature

The speed of sound in air varies with temperature and is given by:

v = 331 + 0.6 * T

Where T is in Celsius. The speed is also related to wavelength and frequency by:

v = λ * f

Given the wavelength λ = 0.80364 m and frequency f = 440 Hz, the wave speed:

v = 0.80364 * 440 ≈ 353.61 m/s

Solving for temperature T:

T = (v - 331) / 0.6 = (353.61 - 331) / 0.6 ≈ 22.68 / 0.6 ≈ 37.8°C

The current air temperature is approximately 37.8°C.

Objects in Oil and Water

When an object with known volume and mass is submerged in layered liquids of different densities, its behavior depends on whether its density is higher or lower than that of the liquids.

Given:

  • Object volume V (not specified, so let’s assume V = V_obj)
  • Mass m = 0.0768 kg
  • Object density ρ_obj = m / V

In layered water and oil systems, water density ≈ 1000 kg/m³, oil density varies around 800-900 kg/m³. Since oil sits on top of water, the system's stable state depends on densities:

  1. If ρ_obj > density of water/oil mixture, the object sinks.
  2. If ρ_obj

Without specific volume or densities, a general conclusion is that if the object's density exceeds that of oil or water, it sinks; if lower, it floats.

Flow Rate Changes in Pipe

Given initial diameter d1=0.35 m, velocity v1=2.2 m/s; larger diameter d2=0.84 m. Using the continuity equation:

A1  v1 = A2  v2

Area is:

A = π * d² / 4

Calculate A1 and A2:

A1 = π * (0.35)^2 / 4 ≈ 0.0954 m²


A2 = π * (0.84)^2 / 4 ≈ 0.554 m²

V2:

v2 = (A1  v1) / A2 ≈ (0.0954  2.2) / 0.554 ≈ 0.418 / 0.554 ≈ 0.755 m/s

The velocity of water in the larger pipe is approximately 0.755 m/s.

Water Jet Pressure and Velocity

The Bernoulli equation relates pressure and velocity:

P + ½ ρ v² + ρ g h = constant

Assuming the height difference is maintained at h=1.8 m, and neglecting elevation change, the velocity v in the jet is:

v = sqrt(2 * (P / ρ))

Given the pressure from the jet, using the relation P = ρ g h, solve for v:

v = sqrt(2 g h) = sqrt(2  9.81  1.8) ≈ sqrt(35.3) ≈ 5.94 m/s

The water in the jet must move at approximately 5.94 m/s to sustain a 1.8 m column height.

Ship Cargo Capacity

Given the ship’s volume V_ship and mass m_ship, the maximum cargo weight before sinking is determined by the buoyant force:

Maximum cargo mass = Displaced water volume × water density - ship’s own mass

Assuming the ship's volume is V and using water density of 1000 kg/m³:

Maximum cargo = (V - m_ship/ρ_water)

Specific numerical results depend on provided values.

References

  • FASB Accounting Standards Codification. (n.d.). ASC 958-605 - Not-for-Profit Entities: Revenue Recognition. Retrieved from https://asc.fasb.org
  • FASB ASC 845-10. (n.d.). Accounting for Transfers of Assets. Retrieved from https://asc.fasb.org
  • Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.
  • Halliday, D., Resnick, R., & Walker, J. (2014). Fundamentals of Physics. Wiley.
  • Holman, J. P. (2010). Experimental methods for engineers. McGraw-Hill.
  • Tipler, P. A., & Mosca, G. (2008). Physics for Scientists and Engineers. W. H. Freeman.
  • Vogel, S. (1994). Life in Moving Fluids: The Physical Biology of Flow. Princeton University Press.
  • Beer, F. P., & Johnston, E. R. (2014). Vector Mechanics for Engineers. McGraw-Hill.
  • Burden, R. L., & Faires, J. D. (2011). Numerical Analysis. Brooks/Cole.
  • American Society of Civil Engineers. (2017). Fluid Mechanics. ASCE Press.

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