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k0ka [10]
3 years ago
11

When you float an ice cube in water, you notice that 90% of it is submerged beneath the surface. Now suppose you put the same ic

e cube in a glass of some liquid whose density is less than that of water.
How much of the ice cube will be submerged below the surface of this liquid?
Physics
1 answer:
Kruka [31]3 years ago
8 0

Answer:

more than 90%

Explanation:

In the unknown liquid the buoyant force and weight relation is

V\rho_lg=mg\\\Rightarrow V\rho_l=m

It can be seen that if the density decreases the buoyant force decreases.

If the object is already 90% submerged in water then, if the other liquid has density less than that of water the object will be submerged more than 90%.

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an elevator stopped on the third floor a basketball shot into a hoop a sled sliding down a snowy hill a tow truck pulling a car out of a ditch

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Why does a geostationary satellite must orbit around Earth's equator, rather than in some other orbit (such as around the poles)
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A satellite on non-equatorial orbit would show daily motion even if its period is exactly 1 sidereal day.

Explanation:

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Mathew throws a ball straight up in the air. it rises for a period of time and then begins to drop. At which points in the balls
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3 years ago
Three point charges are placed on the y-axis: a charge q at y=a, a charge –2q at the origin, and a charge q at y= –a. Such an ar
den301095 [7]

Answer:

electric field   Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

Explanation:

The electric field is a vector, so it must be added as vectors, in this problem both the charges and the calculation point are on the same x-axis so we can work in a single dimension, remembering that the test charge is always positive whereby the direction of the field will depend on the load under analysis, if the field is positive, if the field is negative.

 a) Let's write the electric field for each charge and the total field

       E = k q /r

With k the Coulomb constant, q the charge and r the distance of the charge to the test point

       Et = E1 + E2 + E3

       E1 = k q / (x-a)²

       E2 = k (-2q) / x²  

       E3 = k q / (x + a)²

       Et = kq [1 / (x-a)² -2 / x² + 1 / (x+a)²]

The direction of the field is along the x axis

b) To use a binomial expansion we must have an expression the form (1-x)⁻ⁿ  where x << 1, for this we take factor like x from all the equations

       Et = kq/ x² [1 / (1-a/x)² - 2 + 1 / (1+a/x)²]

We use binomial expansion

     (1+x)⁻² = 1 -nx + n (n-1) 2! x² +… x << 1

     (1-x)⁻² = 1 +nx + n (n-1) 2! x² + ...

They replace in the total field and leaving only the first terms

       

   Et =kq/x² [-2 +(1 +2 a/x + 2 (2-1)/2 (a/x)² +…) + (1 -2 a/x + 2(2-1) /2 (a/x)² +.) ]

   Et = kq/x² [a²/x² + a²/x²2] = kq /x² [2 a²/x²]

Et = k q 2a²/x⁴

point charge

Et = k q 1/x²

Dipole

E = k q a/x³

3 0
3 years ago
Faraday's Law states that the negative of the time rate of change of the flux of the magnetic field through a surface is equal t
MrRa [10]

Answer:

(C). The line integral of the magnetic field around a closed loop

Explanation:

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux.

This can be written mathematically as;

EMF = -\frac{\delta \phi _B}{\delta t}

(\frac{\delta \phi _B}{\delta t} ) is the rate of change of the magnetic flux through a surface bounded by the loop.

ΔФ = BA

where;

ΔФ is change in flux

B is the magnetic field

A is the area of the loop

Thus, according to Faraday's law of electric generators

∫BdL = \frac{\delta \phi _B}{\delta t} = EMF

Therefore, the line integral of the magnetic field around a closed loop is equal to the negative of the rate of change of the magnetic flux through the area enclosed by the loop.

The correct option is "C"

(C). The line integral of the magnetic field around a closed loop

8 0
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