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Zarrin [17]
2 years ago
12

while test flying the new f-22 raptor fighter jet, test pilot bob put the plane in a horizontal loop while flying at the speed o

f sound (343 m/s). what is the minimum radius that the plane can fly if bobs centripetal acceleration cannot exceed 5g (5 times the acceleration due to gravity)?
Physics
1 answer:
sveticcg [70]2 years ago
8 0

Given

v = 343 m/s

ac = 5g

ac = 5*9.8 m/s^2

ac = 49 m/s^2

where,

v: velocity

ac = centripetal aceleration

Procedure

We call the acceleration of an object moving in uniform circular motion—resulting from a net external force—the centripetal acceleration ac; centripetal means “toward the center” or “center seeking”.

Formula

\begin{gathered} a_c=\frac{v^2}{r} \\ r=\frac{v^2}{a_c} \\ r=\frac{(343m/s)^2}{49m/s^2} \\ r=2401\text{ m} \end{gathered}

The minimum radius not to exceed the centripetal acceleration is 2401 m.

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3 years ago
In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi
Ronch [10]

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

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Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

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6 0
4 years ago
A hamster of mass 0.139 kg gets onto his 20.8−cm-diameter exercise wheel and runs along inside the wheel for 0.823 s until its f
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Explanation:

Given that,

Mass of the hamster, m = 0.139 kg

Diameter of the wheel, d = 20.8 cm

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Frequency of the wheel, f = 1 Hz

Time, t = 0.823 s

(a) There exist a relationship between the tangential and angular acceleration of the wheel. It is given by :

a=\alpha \times r

Since, \alpha =\dfrac{\omega}{t}

a=\dfrac{\omega}{t} \times r

a=\dfrac{2\pi f}{t} \times r

a=\dfrac{2\pi \times 1}{ 0.823} \times 10.4\times 10^{-2}

a=0.793\ m/s^2

(b) In radial direction, applying the second law of motion as :

N-mg=ma

a is the radial acceleration, a=\dfrac{v^2}{r}

N=mg+ma

N=mg+m(\dfrac{v^2}{r})

N=mg+m(\dfrac{(r\omega)^2}{r})

N=mg+m\omega^2 r

N=m(g+\omega^2 r)

N=m(g+(2\pi f)^2 r)

N=0.139\times (9.8+(2\pi 1)^2\times 10.4\times 10^{-2})

N=1.93\ N

Hence, this is the required solution.

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