Ask question

Ask question

All categories

4 years ago

11

At 30.0 m below the surface of the sea (density = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air bubble h

aving a volume of 0.95 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble just before it breaks the surface?

1 answer:

stira [4]4 years ago

3 0

Answer:

The volume is $V_a = 1.510 *10^{-5} m^3$

Explanation:

From the question we are told that

The depth below the see is $d_1 = 30.0 \ m$

The density of the sea is $\rho_s = 1025 \ kg /m^3$

The temperature at this level is $T_d = 5.00 ^oC = 278 \ K$

The volume of the air bubble at this depth is $V_d = 0.95 \ cm^3 = 0.95 *0^{-6}\ m$

The temperature at the surface is $T_a = 20^oC =293\ K$

Generally the pressure at the given depth is mathematically evaluated as

$P_d = P_o + \rho_s * g * d$

Where $P_o$ is the atmospheric pressure with a constant value

$P_o = 1.013 *10^{5} \ Pa$

substituting values

$P_d = 1.013 * 10^{5} * + (1025 * 9.8 * 30 )$

$P_d = 4.02650 * 10^{5} \ Pa$

According to the combined gas law

$\frac{P_a * V_a }{T_a } = \frac{P_d * V_d }{T_d }$

=> $V_a = \frac{4.026650 *10^{5} * 0.95 *10^{-6} * 293 }{278 * 1.013*10^{5} }$

=> $V_a = 1.510 *10^{-5} m^3$

You might be interested in

2. It is to bring down the foot forcibly to the f a heavy step) with or without weight. 2 Ormain lateral position - This is done

irga5000 [103]

Im confused bc im not sure if this is a full question

5 0

3 years ago

What are the chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by livin

spin [16.1K]

Answer:

<u>Very low</u>

<u>Explanation:</u>

It is fair to say to a reasonable extent that there are very low chances that radio transmissions from Earth or messages sent on distant space probes will ever be received by living beings .

Bear in mind that for years some scientists have believed without any substantial evidence that there are other living beings in distant space.

4 0

3 years ago

Calculate the angle of refraction of 30.0° light shines from water into ice. The indices of refraction for water and ice are 1.3

WITCHER [35]

Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

refraction , with respect to light or other waves passing through two different substances or media, such as glass, water or air is given by Snell's Law.

Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

Cross multiplying, we have;

n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

Ø2° = Sin–¹(0.8906)

Ø2° = 62.95° ≈ 63°

4 0

3 years ago

A uniform thin wire is bent into a quarter-circle of radius a = 20.0 cm, and placed in the first quadrant. Determine the coordin

Mashcka [7]

Answer:

$r_{cm}=[12.73,12.73]cm$

Explanation:

The general equation to calculate the center of mass is:

$r_{cm}=1/M*\int\limits {r} \, dm$

Any differential of mass can be calculated as:

$dm = \lambda*a*d\theta$ Where "a" is the radius of the circle and λ is the linear density of the wire.

The linear density is given by:

$\lambda=M/L=M/(a*\pi/2)=\frac{2M}{a\pi}$

So, the differential of mass is:

$dm = \frac{2M}{a\pi}*a*d\theta$

$dm = \frac{2M}{\pi}*d\theta$

Now we proceed to calculate X and Y coordinates of the center of mass separately:

$X_{cm}=1/M*\int\limits^{\pi/2}_0 {a*cos\theta*2M/\pi} \, d\theta$

$Y_{cm}=1/M*\int\limits^{\pi/2}_0 {a*sin\theta*2M/\pi} \, d\theta$

Solving both integrals, we get:

$X_{cm}=2*a/\pi=12.73cm$

$Y_{cm}=2*a/\pi=12.73cm$

Therefore, the position of the center of mass is:

$r_{cm}=[12.73,12.73]cm$

5 0

3 years ago

Use cylindrical coordinates. Evaluate E (x − y) dV, where E is the solid that lies between the cylinders x2 + y2 = 1 and x2 + y2

S_A_V [24]

Answer:hchv nhgjj

Explanation:

ggchh

3 0

3 years ago