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3 years ago

Which is most likely a compound?

Physics

1 answer:

Leya [2.2K]3 years ago

4 0

Answer:

I think you forgot to post the sentences

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A sound source of frequency f moves with constant velocity (less than the speed of sound) through a medium that is at rest. A st

ANTONII [103]

Answer:

The correct answer is d

Explanation:

In this exercise they ask us which statement is correct, for this we plan the solution of the problem, this is a Doppler effect problem, it is the frequency change due to the relative speed between the emitter and the receiver of sound.

The expression for the Doppler effect of a moving source is

f ’= (v / (v- + v_s) f

From this expression we see that if the speed the sound source is different from zero feels a change in the frequency.

The correct answer is d

7 0

3 years ago

Until a train is a safe distance from the station it must travel at 5 m/s. Once the train is on open track it can speed up to

Evgesh-ka [11]

Answer: 45x8=360 :)

Explanation:

8 0

3 years ago

How long is the racetrack if it takes a racecar 3.4 s travelling at 75 m/s to finish?

RSB [31]

Answer:

255 metres

Explanation:

The answer to thi question is actually quite simple. Since the car goes for 3.4 seconds, and it goes 75 metres every second, the answer is just 3.4 multiplied by 75 Metres.

5 0

3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

4 years ago

Electric current is.........

BigorU [14]

An electric current is an orderly, directed movement of electrically charged particles

4 0

3 years ago