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frutty [35]
3 years ago
12

Galactic Alliance Junior Mission Officer (GAJMO) Bundit Nermalloy is predicting the kinetic energy of a supply spacecraft, which

is being moved in one dimension in the tractor beam of the ship named the Jadarian-Ruby, to ensure that the supply spacecraft doesn’t damage the spaceport to which it is being delivered. GAJMO Nermalloy has been instructed to deliver the supply spacecraft with a kinetic energy less than 1010J (where 1J=1N⋅m). GAJMO Nermalloy knows that the change in kinetic energy of an object moving in one dimension is equal to the net work performed on it, where net work is the integral of the component of net force in the direction of motion with respect to the position of the of the object. That is: KE2−KE1=∫x2x1F(x)dx. The net force exerted by the tractor beam is supposed to be constant, F0=−3.5×106N, but due to improper maintenance of the Jadarian-Ruby, the actual force exerted by the tractor beam as a function of position x is given by F(x)=αx3+β, where α=6.1×10−9N/m3 and β=−4.1×106N. Assume the supply spacecraft had an initial kinetic energy of KE1=2.7×1011J and that the tractor beam force is applied on the spacecraft over a distance of 7.5×104m away.from its beginning position at x=0.0m.
Physics
1 answer:
antiseptic1488 [7]3 years ago
5 0

Answer:

the ship's energy is greater than this and the crew member does not meet the requirement

Explanation:

In this exercise to calculate kinetic energy or final ship speed in the supply hangar let's use the relationship

                W =∫ F dx = ΔK

                 

Let's replace

             

          ∫ (α x³ + β) dx = ΔK

         α x⁴ / 4 + β x = ΔK

           

Let's look for the maximum distance for which the variation of the energy percent is 10¹⁰ J

         x (α x³ + β) = K_{f} - K₀

          K_{f}  = K₀ + x (α x³ + β)

Assuming that the low limit is x = 0, measured from the cargo hangar

     

Let's calculate

        K_{f}  = 2.7 10¹¹ + 7.5 10⁴ (6.1 10⁻⁹ (7.5 10⁴) 3 -4.1 10⁶)

        Kf = 2.7 10¹¹ + 7.5 10⁴ (2.57 10⁶ - 4.1 10⁶)

        Kf = 2.7 10¹¹ - 1.1475 10¹¹

        Kf = 1.55 10¹¹ J

In the problem it indicates that the maximum energy must be 10¹⁰ J, so the ship's energy is greater than this and the crew member does not meet the requirement

We evaluate the kinetic energy if the System is well calibrated

                W = x F₀ = K_{f} –K₀

                K_{f} = K₀ + x F₀

We calculate

              K_{f} = 2.7 10¹¹ -7.5 10⁴ 3.5 10⁶

               K_{f} = (2.7 -2.625) 10¹¹

              K_{f} = 7.5 10⁹ J

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R_{Al}=2.65(10)^{-8}m\Omega\frac{12m}{0.000004908m^{2}}     (5)

R_{Al}=0.0647\Omega     (6)  Resistance of aluminium wire

<u>Copper wire:</u>

R_{Cu}=1.68(10)^{-8}m\Omega\frac{30m}{0.00000201m^{2}}     (6)

R_{Cu}=0.250\Omega     (7)  Resistance of copper wire

At this point we are able to calculate the  ratio of the resistance of both wires:

Ratio=\frac{R_{Al}}{R_{Cu}}   (8)

\frac{R_{Al}}{R_{Cu}}=\frac{0.0647\Omega}{0.250\Omega}   (9)

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