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tankabanditka [31]
3 years ago
5

You are sitting in a commercial jetliner. It is on the runway and ready to take off. Before it does, clever soul that you are, y

ou hang a .5 meter long string from the ceiling just overhead and attach to its free end a weight whose mass is m = .05 kg. The jet begins its acceleration. As it does, the string and mass swing toward you until the string comes into equilibrium at a constant angle of θ = 26o with the vertical.
a.) What is the jet's acceleration?

b.) Once the jet gains altitude and proceeds with a constant velocity, what will the string and mass be doing (that is, will the 26o angle have changed)? Explain

Physics
1 answer:
wariber [46]3 years ago
4 0

Answer:

a.=4.78 m/s2

b. the thread will remain vertical

Explanation:

a.

For jet acceleration as you know,

Weight due to gravity given as

F= w =mg

and body movement with acceleration as

F= ma

tan\theta=\frac{ma}{mg}

tan\theta=\frac{a}{g}

a=gtan\theta

a=9.81\times tan 26^{0}

a=4.78\frac{m}{s^{2} }

Its acceleration of plane

b.

For planes attaining constant velocity

a = 0  and tan∅ = 0

so Ф = 0

hence the thread will be remain vertical

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If the force of gravity suddenly stopped acting on the planets, they would
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Answer:c

Explanation:

If the Force of gravity suddenly stops acting on Planets then Planets would continue to move straight in the initial direction.

Gravity constantly acts on Planets to change their trajectory each instant thus in absence of it if a planet is moving in a circular path then it would follow the path of the tangent to the circular path as gravity force is absent to change its trajectory.        

4 0
3 years ago
In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th
Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion  (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x  (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

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