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3 years ago

I MARK BRAINLIEST, PLEASE ANSWER ASAP!!! PHYSICS 20 POINTS!!

Physics

2 answers:

JulsSmile [24]3 years ago

6 0

A and c you got thissssss

irina [24]3 years ago

5 0

The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!

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Một nguồn O phát sóng cơ dao động theo phương trình u = 4cos(4pi t - pi/4). Tốc độ truyền sóng là 1m/s.

brilliants [131]

Answer:

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5 0

3 years ago

To do your homework correctly you should (5 points)

pychu [463]

Answer:

C according to me

Explanation:

ITS HARD

4 0

3 years ago

Read 2 more answers

A 67 kg soccer player uses 5100 kJ of energy during a 2.0 h match. What is the

Oksanka [162]

The average power produced by the soccer player is 710 Watts.

Given the data in the question;

Mass of the soccer player; $m = 67kg$

Energy used by the soccer player; $E = 5100KJ = 5100000J$

Time; $t = 2.0h = 7200s$

Power; $P =\ ?$

Power is simply the amount of energy converted or transferred per unit time. It is expressed as:

$Power = \frac{Energy\ converted }{time}$

We substitute our given values into the equation

$Power = \frac{5100000J}{7200s}\\\\Power = 708.33J/s \\\\Power = 710J/s \ \ \ \ \ [ 2\ Significant\ Figures]\\\\Power = 710W$

Therefore, the average power produced by the soccer player is 710 Watts.

Learn more: brainly.com/question/20953664

8 0

2 years ago

A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b

jonny [76]

Answer:

Part a)

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

$P_o = 0.162 Pa$

Explanation:

Part a)

Level of sound = 75 dB

now we know that

$L = 10 Log\frac{I}{I_0}$

here we know that

$I_0 = 10^{-12} W/m^2$

now we have

$75 = 10 Log(\frac{I}{10^{-12}})$

$I = 3.16 \times 10^{-5} W/m^2$

Part b)

Intensity of sound wave is given as

$I = \frac{1}{2}\rho A^2\omega^2 c$

here we know that

$A = \frac{P_o}{Bk}$

so we have

$I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c$

$I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}$

now we know

$\rho = 1.2 kg/m^3$

$c = 340 m/s$

$B = 1.4 \times 10^5 Pa$

now we have

$3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}$

$P_o = 0.162 Pa$

5 0

3 years ago

A sound wave travels in air toward the surface of a freshwater lake and enters into the water. The frequency of the sound does n

Shkiper50 [21]

Answer:

Explanation:

velocity of sound in air at 20⁰C is 343 m /s

velocity of sound in water at 20⁰C is 1481 m /s

The wavelength of the sound is 2.86 m in the air so its frequency

= 343 / 2.86 = 119.93 .

This frequency of 119.93 will remain unchanged in water .

wavelength in water = velocity in water / frequency

= 1481 / 119.93

= 12. 35 m .

7 0

3 years ago