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Marianna [84]
3 years ago
8

If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above a

tmospheric pressure to compress her bones by 0.130 % of their original volume?
Physics
1 answer:
Sever21 [200]3 years ago
8 0

This is an incomplete question, here is a complete question.

The bulk modulus for bone is 15.0 (GPa). If a diver-in-training is put into a pressurized suit, by how much would the pressure have to be raised (in atmospheres) above atmospheric pressure to compress her bones by 0.130% of their original volume?

Answer : The change in pressure will be, 1.95\times 10^7Pa

Explanation : Given,

Bulk modulus = 15.0GPa=15.0\times 10^9Pa

Change in volume = 0.130 % of original volume

Let the original volume be, V

So, Change in volume = \frac{0.130}{100}\times V

Formula used for change in pressure is:

\Delta P=\beta \frac{\Delta V}{V}

Now put all the given values in this formula, we get:

\Delta P=(15.0\times 10^9Pa)\times \frac{(\frac{0.130}{100}\times V)}{V}

\Delta P=1.95\times 10^7Pa

Thus, the change in pressure will be, 1.95\times 10^7Pa

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Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C
miskamm [114]
<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

\boxed{q_{B}=+2q}

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential V, which is <em>the energy per unit charge, </em>so the potential V at any point in an electric field with a test charge q_{0} at that point is:

V=\frac{U}{q_{0}}

The potential V due to a single point charge q is:

V=k\frac{q}{r}

Where k is an electric constant, q is value of point charge and r is  the distance from point charge to  where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius r. Before the sphere A and B touches we have:

V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r

When they touches each other the potential is the same, so:

V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}

Therefore:

(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}

So after A and B touch and are separated the charge on sphere B is:

\boxed{q_{B}=+2q}

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

\boxed{q_{C}=+1.5q}

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

q_{A}=q_{B}=+2q

Second:  C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here q_{A}=+2q and C carries no net charge or q_{C}=0. Also, r_{A}=r_{C}=r

V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

Applying the same concept as the previous problem when sphere touches we have:

k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}

For the principle of conservation of charge:

q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q

Finally, from the third step:

Here q_{B}=+2q \ and \ q_{C}=+q. Also, r_{B}=r_{C}=r

V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}

When sphere touches we have:

k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}

For the principle of conservation of charge:

q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q

So the charge that ends up on sphere C is:

q_{C}=+1.5q

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

+4q

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

+4q

8 0
3 years ago
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