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jarptica [38.1K]
3 years ago
12

At an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s. What is the angle of the shock cone

Physics
1 answer:
Ber [7]3 years ago
5 0

Answer:

55 degrees

Explanation:

Given that an air show a jet flies at speed 1500 km/h on a day when the speed of sound is 342 m/s.

From the question above, we can get the below parameters

Object speed (V) = 1500 km/h

Sound speed ( v) = 342 m/s

Convert km/h to m/s

(1500 × 1000)/3600

Jet speed V = 416.67 m/s

Let's first calculate the mash number M.

M = V/v

M = 416.67 / 342

M = 1.2183

Formula for the angle of the shock cone is reciprocal of mash number. That is,

Sin Ø = 1 / M

Sin Ø = 1 / 1.2183

Sin Ø = 0.8208

Ø = sin^-1(0.8208)

Ø = 55 degree

Therefore, the angle of the shock cone is approximately 55 degrees

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A cyclist has a constant speed of 12 m/s. What is the magnitude of the displacement of the cyclist after 18 seconds?
Katyanochek1 [597]

Answer:

216 m

Explanation:

Assuming a straight line:

Δx = vt

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What is meant by the phrase "a consistent method of measurement"?
julia-pushkina [17]
A measurement that will always give the same answer.
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An object is dropped from a bridge. A second object is thrown downwards 1.00 s later. They both reach the water 20.0 m below at
Deffense [45]

Answer:

v_{o}=-14.60m/s

Explanation:

<u>Kinematics equation for first Object:</u>

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

v_{o}=0m/s       The initial velocity is zero

y_{o}=20m

it reach the water at in instant, t1, y(t)=0:

0=y_{o}-1/2*g*t_{1}^{2}

t_{1}=\sqrt{2y_{o}/g}=\sqrt{2*20/9.81}=2.02s

<u>Kinematics equation for the second Object:</u>

The initial velocity is zero

y(t)=y_{o}+v_{o}t-1/2*g*t^{2}

but:

y_{o}=20m

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s

0=y_{o}+v_{o}t_{2}-1/2*g*t_{2}^{2}

v_{o}=1/(t_{2})*(1/2*g*t_{2}^{2}-y_{o})=(1/1.02)*(1/2*9.81*1.02^{2}-20)=-14.60m/s

The velocity is negative, because the object is thrown downwards.

6 0
3 years ago
A magnetic field is perpendicular to the plane of a single-turn circular coil. The magnitude of the field is changing, so that a
Reika [66]

Answer:

2.62A

Explanation:

Given

V = 0.43 V

I = 3.1 A

Then, V = IR, R = V/I

R = 0.43/3.1

R = 0.14 Ω

The induced emf = dB/dt * A

So that, dB/dt = emf/A

Since dB/dt is constant then Emf/A(circle) = Emf/A square

So Emf (square)/Emf (circle) = A square / A circle

A circle = πr². The perimeter of the square is 2πr which also is the circumference of the square.

Since the perimeter is 2πr, then each side would be πr/2. Thus, the area of the square would be, (πr/2)² = π²r²/4

So A square/Acircle = (π²r²/4) / πr² = π/4 = 0.79

this means that, emf square = emf circle * 0.79

emf square = 0.43*0.79 = 0.34V

I = V/R

I = 0.34/0.13

I = 2.62A

3 0
3 years ago
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