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lana [24]
2 years ago
9

5. Measure: With the lights on, click Pause. Turn on Show rulers. A. The wavelength of a longitudinal wave is equal to the dista

nce between two successive compressions (or rarefactions). What is this distance? B. How does this compare to the wavelength of the comparable transverse wave? (See your answer to question 2C.)
Physics
1 answer:
Marysya12 [62]2 years ago
6 0

Explanation:

A) The distance between the two successive compressions (or rarefactions) is actually called the wavelength of the longitudinal waves.

B) Wavelengths of longitudinal and transverse waves are comparable in the fact that in a transverse wave, the particles move perpendicular to the direction the wave travels whereas in a longitudinal wave the particles are displaced along the direction to the direction the wave travels

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What is the concentration of molecular oxygen (O2) in mol/L on a June day in Toronto when atmospheric pressure is 1.0 atm and th
saveliy_v [14]

Answer:

The concentration of mole evil at oxygen on that day is 0.00858 mol/L

Explanation:

Here, we want to calculate the concentration of molecular oxygen

The pressure on that day is 1.0 atm

Since oxygen is at a concentration of 21%, the pressure of oxygen will be 21/100 * 1 = 0.21 atm

Now let’s calculate the concentration;

From Ideal gas law;

PV = nRT

This can be written as;

P/RT = n/V

The term n/V refers to concentration;

Let’s make substitutions now;

P = pressure = 0.21 atm

R = molar gas constant = 0.0821 L•atm/mol•k

T = temperature = 25 = 25 + 273.15 = 298.15 K

Substituting these values, we have;

n/V = C = 0.21/(0.0821 * 298.15) = 0.00858 mol/L

6 0
2 years ago
The human heart is a powerful and extremely reliable pump. Each day it takes in and discharges about 7500 L of blood. Assume tha
gregori [183]

The answers are:

a) Work=125,923.61J

b) Power=1.46watt

Why?

It seems that you forgot to write the questions of the problem, however, in order to help you, I will try to complete it.

The questions are:

a) How much work does the heart do in a day?

b) What is its power output in watts?

So, solving we have:

We need to convert from liter to cubic meters in order to use the given information, so:

1L=0.001m^{3}\\\\7500L*\frac{0.001m^{3} }{1L}=7.5m^{3}

Also, we need to find the mass given the density of the blood.

1050}\frac{kg}{m^{3}}*7.5m^{3}=7875kg

Now, calculating how much work does the heart do in a day, we have:

Work=Fd=mgh\\\\Work=7875kg*9.81\frac{m}{s^{2}}*1.63m=125,923.61J

Then, calculating what is the power output and its horsepower, we have:

Power=\frac{Work}{time}\\\\Power=\frac{125,923.61J}{86,400s}=1.46watt

Have a nice day!

7 0
2 years ago
Imagine you’re driving along a road and you approach a bridge. You notice a sign that reads, “Bridge freezes before road.” Why d
nydimaria [60]

<u>Answer:</u>

<h3>During wet and freezing temperatures, ice is able to form at a faster pace on bridges because freezing winds blow from above and below and both sides of the bridge, causing heat to quickly escape. The road freezes slower because it is merely losing heat through its surface.</h3>

<u>Sources:</u>

-- https://intblog.onspot.com/en-us/why-do-bridges-become-icy-before-roads

and

-- https://www.accuweather.com/en/accuweather-ready/why-bridges-freeze-before-roads/687262

I hope this helps you! ^^

6 0
2 years ago
When explaining chemical reactions to a friend, Brianna models a reaction by taking several colors of modeling clay and making a
Drupady [299]

Answer: synthesis

Explanation:

5 0
2 years ago
Read 2 more answers
Inna Hurry is traveling at 6.8 m/s, when she realizes she is late for an appointment. She accelerates at 4.5 m/s^2 for 3.2 s. Wh
Alborosie

Answer:

1) v = 21.2 m/s

2) S = 63.33 m

3) s = 61.257 m

4) Deceleration, a = -4.32 m/s²

Explanation:

1) Given,

The initial velocity of Inna, u = 6.8 m/s

The acceleration of Inna, a = 4.5 m/s²

The time of travel, t = 3.2 s

Using the first equation of motion, the final velocity is

                v = u + at

                   = 6.8 + 4.5 x 3.2

                   = 21.2 m/s

The final velocity of Inna is, v = 21.2 m/s

2) Given,

The initial velocity of Lisa, u = 12 m/s

The final velocity of Lisa, v = 26 m/s

The acceleration of Lisa, a = 4.2 m/s²

Using the III equations of motion, the displacement is

                          v² = u² +2aS

                         S = (v² - u²) / 2a

                            = (26² -12²) / 2 x 4.2

                            = 63.33 m

The distance Lisa traveled, S = 63.33 m

3) Given,

The initial velocity of Ed, u = 38.2 m/s

The deceleration of Ed, d = - 8.6 m/s²

The time of travel, t = 2.1 s

Using the II equations of motion, the displacement is

                        s = ut + 1/2 at²

                           =38.2 x 2.1 + 0.5 x(-8.6) x 2.1²

                           = 61.257 m

Therefore, the distance traveled by Ed, s = 61.257 m

4) Given,

The initial velocity of the car, u = 24.2 m/s

The final velocity of the car, v = 11.9 m/s

The time taken by the car is, t = 2.85 s

Using the first equations of motion,

                         v = u + at

∴                        a = (v - u) / t

                            = (11.9 - 24.2) / 2.85

                            = -4.32 m/s²

Hence, the deceleration of the car, a = = -4.32 m/s²

5 0
2 years ago
Read 2 more answers
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