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faust18 [17]
3 years ago
11

Lucy is cruising through space in her new spaceship. As she coasts along, a tiny spacebug drifts into her path and bounces off t

he window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.
Physics
1 answer:
Elden [556K]3 years ago
4 0

Your question is not complete, please let me assume this to be your complete question:

Lucy is cruising through space in her new spaceship. As she coasts along, a spacebug drifts into her path and bounces off the window. Consider several statements concerning this scenario. Evaluate each statement according to the law of momentum conservation and match it to the appropriate category.

Below are several statements concerning this scenario. Evaluate each statement and decide if it is true, false, or undetermined by the principle of momentum conservation.

i) The total change in momentum for this interaction is zero.

ii) The change in the space bug's momentum is greater than the change in the spaceship's momentum.

iii) If the space bug had stuck to the spaceship instead of bouncing off, momentum would not have been conserved for this interaction.

Answer:

I is true

II is false

III is undetermined

Explanation:

STATEMENT I : This statement is TRUE because because they was no change in velocity and mass of the spaceship and spacebug, after the collision, as the bug bounced outside the window and the spaceship retains it's velocity. Therefore the total momentum in the system before the collision, is equal to the total momentum of the system after the collision.

Where;

momentum = Mass × velocity

M1V1 = M2V2

Therefore;

M2V2 - M1V1 = 0

STATEMENT II : This statement is FALSE, because the change in the momentum of the spaceship and bug are equal, as the spaceship and bug remains in constant motion after collision. The collision did not have any effect in their velocity nor mass.

Ps = Pb

Ps = Momentum of spaceship

Pb = Momentum of spacebug

STATEMENT III : The statement is UNDETERMINE, because it will depend on the momentum we are considering. If the spaceship is still in constant motion, that means moment of the spaceship is conserved, while that of the spacebug is not conserved.

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A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
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Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

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