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3 years ago

A sound wave has a speed of 345 m/s and a wavelength of 500 meters. Is it infrasonic, sonic, or ultrasonic

Physics

1 answer:

exis [7]3 years ago

4 0

Answer:

Infrasonic.

Explanation:

Given the following data;

Speed = 345 m/s

Wavelength = 500 meters

To find the frequency of the sound wave;

Speed = wavelength * frequency

Frequency = speed/wavelength

Frequency = 345/500

Frequency = 0.69 Hz

Therefore, the sound wave is infrasonic.

Note: Frequency of infrasonic sound = below 20 Hertz.

Frequency of sonic sound = 20 Hz to 20,000 Hertz.

Frequency of ultrasonic sound = above 20,000 Hertz.

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A 69.5-kg person throws a 0.0475-kg snowball forward with a ground speed of 31.5 m/s. A second person, with a mass of 57.5 kg, c

Leno4ka [110]

Answer:

- After throwing the snow, velocity of the thrower is 2.33 m/s

- the velocity of the receiver is 0.026 m/s

Explanation:

Given the data in the question;

Using conservation of momentum,

Initial thrower has a momentum of mv; $m_{total$ v

(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s

Now, When he throws it at 31.5 m/s, these constitutes a momentum of;

(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s

hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s

To get his velocity, we say;

161.94035 = mv

{ he lost weight of the snow ball so, m = 69.5 kg )

161.94035 = 69.5 × v

v = 161.94035 / 69.5

v = 2.33 m/s

Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s

Next is the Receiver;

the receiver will gain momentum of 1.49625 kg.m/s

he has no momentum initially and after he catches the snow ball;

1.49625 kg.m/s = mv

1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v

1.49625 kg.m/s = 57.5475 kg × v

v = ( 1.49625 kg.m/s ) / 57.5475 kg

v = 0.026 m/s

Therefore, the velocity of the receiver is 0.026 m/s

3 0

3 years ago

Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s west

Aleks04 [339]

Explanation:

Given that,

Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.

Taking eastward as positive direction, we have:

$v_B=+5\ m/s$ is the velocity of Bill with respect to Amy (which is stationary)

$v_c=15\ m/s$ is the velocity of Carlos with respect to Amy.

Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is

$v_B=+5\ m/s$

Therefore, Carlos velocity in Bill's reference frame will be

$v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s$

So, the magnitude is 20 m/s and the direction is westward (negative sign).

7 0

3 years ago

The work output of a machine divided by the work input is the ____ of the machine.

s344n2d4d5 [400]

<span>The work output of a machine divided by the work input is the "Efficiency" of the machine.

Hope this helps!</span>

7 0

3 years ago

A 14gram ovarian tumor is treated using a sodium phosphate in which the phosphorus atoms are the radioactive phosphorus 32 isoto

Nata [24]

I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd

6 0

3 years ago

A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

3 years ago