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Angelina_Jolie [31]
3 years ago
8

Some lakes, such as the Great Salt Lake, accumulate soluble minerals such as salt. In those lakes, people find it much easier to

float than when they are in fresh water. Why is this the case?
Physics
2 answers:
olga55 [171]3 years ago
5 0
<span>The water with a lot of dissolved minerals and salts is much denser than fresh water. And with a higher density, that means that less water needs to be displaced in order to float. This difference in density can be the difference between barely floating on your back with your nose barely out of the water versus floating comfortably on your back with your entire face out of the water. For instance, fresh water at 25C has a density of 997 kg/m^3, whereas the water in the Great Salt Lake, depending upon the salinity ranges from 1035 kg/m^3 to 1200 kg/m^3. Or in simpler terms it can be about 4% to 20% denser than fresh water and you'll float 4% to 20% higher in the salt water.</span>
erastova [34]3 years ago
4 0

Answer:

B)  The salt water has a greater density than fresh water.

Explanation:

Took the test

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A 120 g, 8.0-cm-diameter gyroscope is spun at 1000 rpm and allowed to precess. What is the precession period?
dolphi86 [110]

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

I = MR^2

Here,

M = Mass

R = Radius of the hoop

The precession frequency is given as

\Omega = \frac{Mgd}{I\omega}

Here,

M = Mass

g= Acceleration due to gravity

d = Distance of center of mass from pivot

I = Moment of inertia

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Replacing the value for moment of inertia

\Omega= \frac{MgR}{MR^2 \omega}

\Omega = \frac{g}{R\omega}

The value for our angular velocity is not in SI, then

\omega = 1000rpm (\frac{2\pi rad}{1 rev})(\frac{1min}{60s})

\omega = 104.7rad/s

Replacing our values we have that

\Omega = \frac{9.8m/s^2}{(8*10^{-2}m)(104.7rad)}

\Omega = 1.17rad/s

The precession frequency is

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7 0
2 years ago
Two 0.50 g spheres are charged equally and placed 2.5 cm apart. When released, they begin to accelerate at 170 m/s^2 .What is th
vitfil [10]

Answer:

q=7.65*10^{-8}C

Explanation:

Using Newton's second law, we calculate the magnitude of the electric force between the spheres:

F=ma\\F=0.5*10^{-3}kg(170\frac{m}{s^2})\\F=0.085N

The magnitude of the charge in both spheres is the same. So, we calculate the charge, using Coulomb's law:

F=\frac{kq^2}{d^2}\\q=\sqrt\frac{Fd^2}{k}\\q=\sqrt\frac{(0.085N)(2.5*10^{-2}m)^2}{8.99*10^9\frac{N\cdot m^2}{C^2}}\\q=7.65*10^{-8}C

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