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ki77a [65]
3 years ago
10

As SCUBA divers go deeper underwater, the pressure from the weight of all the water above them increases tremendously which comp

resses the gases in their blood. What happens to the volume of gas in their blood as the diver rises quickly to the surface?
Physics
1 answer:
faltersainse [42]3 years ago
5 0

Answer:  The volume of gas expands because of the decrease in pressure as he tries to exit the water body, therefore he must take necessary precaution.

Explanation:

Using Boyle's law which states that the  the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature

ie P1VI=P2V2

A diver absorbs compressed nitrogen gas when  he dives into the water body, As he ascends  out of the water body having less pressure, the volume of nitrogen gas which he absorbs will tend to expand following  Boyle's Law.  Therefore a scuba driver should not rises quickly but slowly  to the surface or else the  expanding nitrogen gas can cause tiny bubbles in his blood and tissue to form together with joints pains and eventually  cause decompression sickness needing medical attention.

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Electromagnetic waves travel at a speed of 3.0 x 108 m/s. A radio station transmits waves
Sophie [7]

Answer:

1.2×10⁶ Hz

Explanation:

Applying,

v = λf............. Equation 1

Where v = speed of wave, λ = wavelength, f = frequency.

make f the subject of the equation

f = v/λ............... Equation 2

From the question,

Given: v = 3.0×10⁸ m/s, λ = 2.5×10² m

Substitute these values into equation 2

f =  3.0×10⁸/2.5×10²

f = 1.2×10⁶ Hz

6 0
3 years ago
2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
3 years ago
A ball is thrown vertically upward, which is the positive direction. A little later, it returns to its point of release. The bal
Aleks [24]

Answer:

The initial velocity of the ball is <u>39.2 m/s in the upward direction.</u>

Explanation:

Given:

Upward direction is positive. So, downward direction is negative.

Tota time the ball remains in air (t) = 8.0 s

Net displacement of the ball (S) = Final position - Initial position = 0 m

Acceleration of the ball is due to gravity. So, a=g=-9.8\ m/s^2(Acting down)

Now, let the initial velocity be 'u' m/s.

From Newton's equation of motion, we have:

S=ut+\frac{1}{2}at^2

Plug in the given values and solve for 'u'. This gives,

0=8u-0.5\times 9.8\times 8^2\\\\8u=4.9\times 64\\\\u=\frac{4.9\times 64}{8}\\\\u=4.9\times 8=39.2\ m/s

Therefore, the initial velocity of the ball is 39.2 m/s in the upward direction.

3 0
4 years ago
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface.
elena-s [515]

Answer:

g=13.42\frac{m}{s^2}

Explanation:

1) Notation and info given

\rho_{center}=13000 \frac{kg}{m^3} represent the density at the center of the planet

\rho_{surface}=2100 \frac{kg}{m^3} represent the densisty at the surface of the planet

r represent the radius

r_{earth}=6.371x10^{6}m represent the radius of the Earth

2) Solution to the problem

So we can use a model to describe the density as function of  the radius

r=0, \rho(0)=\rho_{center}=13000 \frac{kg}{m^3}

r=6.371x10^{6}m, \rho(6.371x10^{6}m)=\rho_{surface}=2100 \frac{kg}{m^3}

So we can create a linear model in the for y=b+mx, where the intercept b=\rho_{center}=13000 \frac{kg}{m^3} and the slope would be given by m=\frac{y_2-y_1}{x_2-x_1}=\frac{\rho_{surface}-\rho_{center}}{r_{earth}-0}

So then our linear model would be

\rho (r)=\rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r

Since the goal for the problem is find the gravitational acceleration we need to begin finding the total mass of the planet, and for this we can use a finite element and spherical coordinates. The volume for the differential element would be dV=r^2 sin\theta d\phi d\theta dr.

And the total mass would be given by the following integral

M=\int \rho (r) dV

Replacing dV we have the following result:

M=\int_{0}^{2\pi}d\phi \int_{0}^{\pi}sin\theta d\theta \int_{0}^{r_{earth}}(r^2 \rho_{center}+\frac{\rho_{surface}-\rho_{center}}{r_{earth}}r)

We can solve the integrals one by one and the final result would be the following

M=4\pi(\frac{r^3_{earth}\rho_{center}}{3}+\frac{r^4_{earth}}{4} \frac{\rho_{surface}-\rho_{center}}{r_{earth}})

Simplyfind this last expression we have:

M=\frac{4\pi\rho_{center}r^3_{earth}}{3}+\pi r^3_{earth}(\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}(\frac{4}{3}\rho_{center}+\rho_{surface}-\rho_{center})

M=\pi r^3_{earth}[\rho_{surface}+\frac{1}{3}\rho_{center}]

And replacing the values we got:

M=\pi (6.371x10^{6}m)^2(\frac{1}{3}13000 \frac{kg}{m^3}+2100 \frac{kg}{m^3})=8.204x10^{24}kg

And now that for any shape the gravitational acceleration is given by:

g=\frac{MG}{r^2_{earth}}=\frac{(6.67408x10^{-11}\frac{m^3}{kgs^2})*8.204x10^{24}kg}{(6371000m)^2}=13.48\frac{m}{s^2}

4 0
3 years ago
A 0.40-μF capacitor is connected to a 5.0-V battery. How much charge is on each plate of the capacitor? Express your answer usin
weqwewe [10]

Answer:

Q=2.0\times10^{-6}C

Explanation:

The equation of charge on the capacitor given its capacitance and a voltage applied to it is Q=CV.

Our capacitance is expressed in <em>microfarads</em>, where micro stands for \times10^{-6}, so in S.I. we sill have:

Q=CV=(0.4\times10^{-6}F)(5V)=0.000002C=2.0\times10^{-6}C

Where the results has been expresed with <em>two significant figures</em>.

7 0
3 years ago
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