Ask question

Ask question

All categories

3 years ago

5

A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is dr

iven toward it at 1/4 the speed of sound. Show that a person sitting in the plane will hear the same frequency of sound from the siren.

1 answer:

melomori [17]3 years ago

4 0

Answer:

See the attachments below.

Explanation:

You might be interested in

What is it called when sediment is dropped and comes to rest?

larisa86 [58]

I think d would be the answer.

3 0

4 years ago

Read 2 more answers

What is the correct path of electrons?

MrRa [10]

Electrons are transferred sequentially between the two photosystems, with photosystem I acting to generate NADPH and photosystem II acting to generate ATP. The pathway of electron flow starts at photosystem II, which is homologous to the photosynthetic reaction center of R. viridis already described.

3 0

3 years ago

How well a student reads can most likely be effected by

Burka [1]

Their cognitive skills and their ability to learn

3 0

3 years ago

5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co

Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop. At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0

3 years ago

Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago