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Bezzdna [24]
3 years ago
6

slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

ane is 980 km/h, and the jet stream it will fly through is presumed to move due east. What is the assumed speed of the jet stream
Physics
1 answer:
11111nata11111 [884]3 years ago
4 0

Answer:

v_{s}=65.2km/h

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

So

t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0

Apply quadratic formula to solve for vs

So

v_{s}=65.2km/h

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1) 256.9 N/m

The force applied to the spring is equal to the weight of the block hanging on the spring:

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And the spring constant can be found by using Hook's law, because we know that the displacement caused by this force is x = 0.29 m:

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2) 1.08 Hz

The angular frequency of oscillation of the spring is given by the formula:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{256.9 N/m}{7.6 kg}}=5.81 rad/s

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3) 2.19 m/s

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t is the time

Substituting t=0.36 s, we find the speed of the block at that time:

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4) 25.6 m/s^2

The maximum acceleration is given by:

a_0 = \omega^2 A

where A is the amplitude of the oscillation.

We can find the amplitude by using the law of conservation of energy: in fact, the kinetic energy at the equilibrium point must be equal to the elastic potential energy at the point of maximum displacement:

K=U\\\frac{1}{2}mv_0^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{mv_0^2}{k}}=\sqrt{\frac{(7.6 kg)(4.4 m/s)^2}{256.9 N/m}}=0.76 m

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a_0 = \omega^2 A=(5.81 rad/s)^2 (0.76 m)=25.6 m/s^2

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