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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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LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But  ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

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3 years ago
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They cause an increase in temp of earths atmosphere or warming by absorbing solar energy. hope this helps
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3 years ago
A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceler
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Answer:

331665750000\ m/s^2

3257806.62409 m/s

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

M = Mass of Sun = 1.989\times 10^{30}\ kg

r = Radius of Star = 20 km

u = Initial velocity = 0

v = Final velocity

s = Displacement = 16 m

a = Acceleration

Gravitational acceleration is given by

g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2

The gravitational acceleration at the surface of such a star is 331665750000\ m/s^2

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s

The velocity of the object would be 3257806.62409 m/s

8 0
3 years ago
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Answer: 0.53m

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Initial velocity given (v₀) = 3.25 m/s

The body is also under the influence of gravity but the acceleration due to gravity will be negative being an upward force (a = -g) and the distance (s) will serve as our maximum height (h)

The equation of motion will.now become

V = v₀² -2gh

Where v = 0 v₀ = 3.25m/s g = 10m/s h = ?

0 = 3.25² - 2(10)h

0 = 10.56 - 20h

-10.56 = -20h

h = 10.56/20

h = 0.53m

Therefore, the maximum height, h (in meters), above the launch point that the basketball will achieve is 0.53m

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Answer: A = square root (2/L)

Explanation: find the attached file for explanation

7 0
3 years ago
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