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JulsSmile [24]
3 years ago
13

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a g

round observer is the sound of the aircraft heard by the ground observer
Physics
1 answer:
Gre4nikov [31]3 years ago
7 0

Answer:

<h3>After 3seconds</h3>

Explanation:

A supersonic aircraft flies at 3 km altitude at a speed of 1000 m/s on a standard day. How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer

Using the formula for calculating speed expressed as;

Speed = Distance/Time

Given;

Distance = 3km = 3000m

Speed = 1000m/s

Required

How long after passing directly above a ground observer is the sound of the aircraft heard by the ground observer (Time)

From the formula;

Time = Distance/speed

Time = 3000/1000

Time = 3seconds

Hence the sound of the aircraft is heard after 3 seconds

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When a constant force acts on an object, what does the object's change in momentum depend upon?
Eva8 [605]
You don’t need time (1) or the force (4) either so just 3,5
8 0
3 years ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
4 years ago
Two uniform solid cylinders, each rotating about its central (longitudinal) axis, have the same mass of 2.88 kg and rotate with
Tanya [424]

Answer:

(a) K_{small}=4839.3J

(b) K_{larger}=17406.4J

Explanation:

Given data

The angular velocity of two cylinders ω=257 rad/s

The mass of the two cylinders m=2.88 kg

The radius of small cylinder r₁=0.319 m

The radius of larger cylinder r₂=0.605 m

For Part (a)

The rotational kinetic energy of the cylinder is given by:

K=\frac{1}{2}Iw^2

Where I is rotational of inertia of solid cylinder about its central axis.

So

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr^2)w^2

Substitute the given values

So

K_{small}=\frac{1}{4}(2.88kg)(0.319)^2(257rad/s)^2 \\K_{small}=4839.3J

For Part (b)

K=\frac{1}{2}Iw^2\\ K=\frac{1}{2}(1/2mr_{2}^2)w^2

Substitute the given values

K_{larger}=\frac{1}{4}mr_{2}^2w^2\\ K_{larger}=\frac{1}{4}(2.88kg)(0.605m)^2(257rad/s)^2\\ K_{larger}=17406.4J

8 0
3 years ago
A 1425 kg truck driving at 13.0 m/s collides elastically with a stationary 1175 kg car. If the car is traveling 14.25 m/s just a
TEA [102]

Answer:

1.25 m/s

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(1425*13)+(1175*0)=(1425*v1f)+(1175*14.25)

18525+0=1425(v1f)+16743.75

1781.25=1425(v1f)

v1f=1.25 m/s

4 0
3 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

\tau=\sigma\cos\alpha\cos\beta

Put the value into the formula

\tau=12\cos60\times\cos35

\tau=4.91\ MPa

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}

Put the value into the formula

\sigma=\dfrac{6.2}{\cos60\cos35}

\sigma=15.13\ MPa

Hence, This is the required solution.

3 0
3 years ago
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