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mixas84 [53]
4 years ago
11

An airplane flying at 115 m/s due east makes a gradual turn following a circular path to fly south. The turn takes 15 seconds to

complete. a. What is the radius of the curve that the plane follows in making the turn? b. What is the magnitude of the centripetal acceleration during the turn?
Physics
1 answer:
bulgar [2K]4 years ago
3 0

To solve this exercise it is necessary to apply the concepts related to Centripetal and Perimeter acceleration of a circle.

The perimeter of a circle is defined by

P = 2\pi r

Where,

r= radius

While centripetal acceleration is defined by

a=\frac{v^2}{r}

Where,

v= velocity

r= radius

PART A)

The distance of a body can be defined based on the speed and the time traveled, that is

x = v*t

For our values the distance is equal to

x = 15*115=1725m

The plane when going to make the turn from east to south makes a quarter of the circumference that is

\frac{P}{4} = \frac{2\pi r}{4}

The same route you take is the distance traveled, that is

x = \frac{P}{4}

x = \frac{2\pi r}{4}

1725 = \frac{2\pi r}{4}

r = 1098.17m

PART B)

With the radius is possible calculate he centripetal acceleration,

a = \frac{v^2}{r}

a = \frac{115^2}{1098.17}

a = 12.04m/s^2

Therefore the radius of the curva that the plane follows in making the turn is 1098.17m with a centripetal acceleration of 12.04m/s^2

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maxonik [38]

Wave length(K) = 1.3*10^{7} rad/m

wavelength = \frac{2n}{k}

wavelength = \frac{2π}{1.37*10^7}

wavelength = 4.584*10^{-7}/m^{-1}

wavelength = 4.58*10^{-7} m

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The distance over which a periodic wave's shape repeats is known as the wavelength in physics. It is a property of both traveling waves and standing waves as well as other spatial wave patterns. It is the distance between two successive corresponding locations of the same phase on the wave, such as two nearby crests, troughs, or zero crossings. The spatial frequency is the reciprocal of wavelength. The Greek letter lambda () is frequently used to represent wavelength. The term wavelength is also occasionally used to refer to modulated waves, their sinusoidal envelopes, or waves created by the interference of several sinusoids.

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5 0
1 year ago
Consider two displacements, one of magnitude 3 m and another of magnitude 4 m. Show how the displacement vectors may be combined
Rufina [12.5K]

Answer:

Explanation:

A = 3m

B = 4 m

let the angle between the two vectors is θ.

the resultant of two vectors is given by

R=\sqrt{A^{2}+B^{2}+2ABCos\theta }

(a) R = 7 m

So, 7=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

49 = 9 + 16 + 24 Cosθ

Cosθ = 1

θ = 0°

Thus, the two vectors are inclined at 0°.

(b) R = 1 m

So, 1=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

1 = 9 + 16 + 24 Cosθ

Cosθ = - 1

θ = 180°

Thus, the two vectors are inclined at 180°.

(c) R = 5 m

So, 5=\sqrt{3^{2}+4^{2}+2\times 3\times 4 Cos\theta }

25 = 9 + 16 + 24 Cosθ

Cosθ = 0

θ = 90°

Thus, the two vectors are inclined at 90°.

3 0
3 years ago
A boy throws a ball upward with a velocity of 4.50 m/s at 60.0o. What is the maximum height reached by the ball?
Cloud [144]

Answer:

3.1m

Explanation:

Since we only care about the y direction we only need to find vy. Once u draw your vector you will realize that vy= 4.5sin60=3.897m/s.

use vf²=v²+2a(y)

At the maximum height the velocity is 0 and since the object is in freefall, a=-g

Plug in all values

0=15.1875-2*9.8(y)

solve for y

-15.1875*2/-9.8=y

y=3.1m

3 0
3 years ago
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Physics part 1 <br><br> I need help answering these
alex41 [277]

Answer:

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6 0
3 years ago
g 1. Water flows through a 30.0 cm diameter water pipe at a speed of 3.00 m/s. All of the water in the pipe flows into a smaller
levacccp [35]

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

A_1v_1 = A_2v_2

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

(0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}

<u>v₂ = 30 m/s</u>

<u></u>

b)

m_1 = \rho A_1v_1t

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\

<u>m₁ = 12600 kg</u>

<u></u>

c)

m_1 = \rho A_1v_1t

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\

<u>m₂ = 12600 kg</u>

7 0
3 years ago
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