What allows a metal to be pounded into a flat shape?

Which of the following requirements must be present to hear a sonic boom?

Lorico [155]

Answer:

D) The velocity of the airplane must be greater than the speed of sound

Explanation:

Sonic boom is a phenomenon that occurs when an airplane travels ay a velocity greater than the speed of sound.

When this condition is met, the sound waves produced by the airplane (mainly from the engine) is not able to travel ahead of the plane, because the plane is moving at a greater speed than the sound itself; as a result, it is only transmitted behind the plane.

In this situation, a shock wave is produced. This shock wave is associated with a very loud "bang", that can be heard clearly even at several kilometers distance (in fact, supersonic planes are normally not allowed to fly over residential areas. In the past, for example the Concorde, was only allowerd to fly at supersonic speed over the ocean).

So the correct answer is

D) The velocity of the airplane must be greater than the speed of sound

6 0

3 years ago

An object executing simple harmonic motion has a maximum speed of 4.3 m/s and a maximum acceleration of 0.65 m/s2 . find (a) the

Alexxx [7]

(1) The position around equilibrium of an object in simple harmonic motion is described by
$x(t) = A \cos (\omega t)$
where
A is the amplitude of the motion
$\omega$ is the angular frequency.

The velocity is the derivative of the position:
$v(t)=-\omega A \sin(\omega t) = -v_0 \sin (\omega t)$
where
$v_0 = \omega A$ is the maximum velocity of the object.

The acceleration is the derivative of the velocity:
$a(t)=- \omega^2 A \cos (\omega t) = -a_0 \cos (\omega t)$
where
$a_0=\omega^2 A$ is the maximum acceleration of the object.

We know from the problem both maximum velocity and maximum acceleration:
$v_0 = \omega A = 4.3 m/s$
$a_0 = \omega^2 A = 0.65 m/s^2$
From the first equation, we get
$A= \frac{4.3 }{\omega}$ (1)
and if we substitute this into the second equation, we find the angular fequency
$\omega=0.15 rad/s$
while the amplitude is (using (1)):
$A=28.7 m$

(b) We found in the previous step that the angular frequency of the motion is
$\omega=0.15 rad/s$
But the angular frequency is related to the period by
$T= \frac{2 \pi}{\omega}$
and so, the period is
$T= \frac{2 \pi}{\omega}= \frac{2 \pi}{0.15 rad/s}=41.9 s$

5 0

4 years ago

Use the factorization to find the values of x for

TiliK225 [7]

Answer:

Nah son

Explanation:

8 0

4 years ago

Read 2 more answers

A sphere has surface area 1.25 m2, emissivity 1.0, and temperature 100.0°C. What is the rate at which it radiates heat into empt

Yuri [45]

The total power emitted by an object via radiation is:
$P=A\epsilon \sigma T^4$
where:
A is the surface of the object (in our problem, $A=1.25 m^2$
$\epsilon$ is the emissivity of the object (in our problem, $\epsilon=1$ )
$\sigma = 5.67 \cdot 10^{-8} W/(m^2 K^4)$ is the Stefan-Boltzmann constant
T is the absolute temperature of the object, which in our case is $T=100^{\circ} C=373 K$

Substituting these values, we find the power emitted by radiation:
$P=(1.25 m^2)(1.0)(5.67 \cdot 10^{-8}W/(m^2K^4)})(373 K)^4=1371 W = 1.4 kW$
So, the correct answer is D.

6 0

4 years ago

A car is moving at constant velocity +48.8 m/s along a straight highway. Suddenly the driver spots a stalled vehicle 378.7 m ahe

Katen [24]

Answer:

a = - 3.14[m/s^2]

Explanation:

To solve this problem we must use the following kinematic equation:

$v_{f} ^{2}= v_{o} ^{2}+2*a*dx\\where:\\v_{f} = final velocity = 0 [m/s]\\v_{o}= initial velocity = 48.8 [m/s]\\ a = acceleration [m/s^2]\\dx = displacement [m]\\$

Now replacing:

$0 = 48.8^{2}+2*a*(378.7) \\-2381.44 = 757.4*a\\a = -3.14[m/s^2]$

The negative sign means the vehicle slows down.

4 0

4 years ago