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andriy [413]
3 years ago
11

Why does ice float on water? A. The temperature of ice is lower than the temperature of water. B. The temperature of water and i

ce are the same. C. The density of ice is less than the density of water. D. The density of ice is greater than the density of water.
Physics
2 answers:
Allushta [10]3 years ago
4 0

It is C, ice is about 9% less dense than water! This is true info

Vesnalui [34]3 years ago
4 0

Answer:

C. The density of ice is less than the density of water.

Explanation:

As we know that when ice floats on the water then we can say that

Buoyancy force on ice will be counterbalanced by the weight of ice block

So here we have

F_b = mg

now we say that for flotation of ice cube the buoyant force must be greater than the force of gravity

\rho_w Vg > \rho_{ice} Vg

so here we can say

\rho_w > \rho_{ice}

so here density of ice will be less than the density of water

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Great Question! I happened to be a physics nerd!

Answer:

C. Two hydrogen nuclei, each with only one proton, fuse to form deuterium, a form of hydrogen with one proton.

MAKE SURE TO SEE EXPLANATION!

Explanation:

In the core of the Sun, or any other main sequence star, there is no single fusion process. Instead, complex sequences of processes occur to make helium nuclei from hydrogen nuclei (i.e. protons). The proton-proton chain provides for the majority of energy generation in stars with masses less than that of the Sun.  One difficulty in creating a helium nucleus (two protons and two neutrons) is that there are only protons to begin with. Some protons must be turned into neutrons in some way. The first step is to combine two protons to form a deuterium nucleus (also known as a deuteron). That's a hefty hydrogen nucleus with one proton and one neutron. Such a proton-proton contact is highly unlikely, and it has never been detected in a laboratory. Fortunately, the Sun's core is incredibly hot and dense, with an incredible number of protons packed inside. Even a low likelihood event will occur every now and again. Along with each deuteron, a positron (an "anti-electron") and a neutrino are created. Because the Sun's core is plasma, there are a lot of free electrons, thus the positron doesn't live long until it and an electron collide and annihilate, resulting in gamma radiation. The deuteron then interacts with a proton to form a helium 3 nucleus. That is a high-probability interaction, and it occurs swiftly. Two helium 3 nuclei join in the third phase to generate a helium 4 ("regular" helium) nucleus and a proton. Branch I of the proton-proton (p-p) chain is responsible for this. Another stage is required because reactions between helium 3 and helium 4 nuclei are possible. There are two conceivable reactions (named Branch II and Branch III), and I'll save you the gory details. It gets much more complicated since theoretical calculations indicate that a reaction between a helium 3 nucleus and a proton is feasible — Branch IV. This reaction has an incredibly low likelihood of occurring, far lower than the Branch I reaction, thus it must be exceedingly rare. The Carbon-Nitrogen-Oxygen (CNO) Cycle is another method for reducing hydrogen to helium. It does not generate much energy in the Sun, but it is the principal energy generation mechanism in larger stars.

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An object is located 50 cm from a converging lens having a focal length of 15 cm. Which of the following is true regarding the i
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Answer:

It is real, inverted, and smaller than the object.

Explanation:

Let's start by using the lens equation to find the location of the image:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where we have:

q = ? is the distance of the image from the lens

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p = 50 cm is the distance of the object from the lens

Solving the equation for q, we find

\frac{1}{q}=\frac{1}{15 cm}-\frac{1}{50 cm}=0.047 cm^{-1}

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The sign of q is positive, so the image is real.

Now let's also write the magnification equation:

h_i = - h_o \frac{q}{p}

where  

h_i, h_o are the size of the image and of the object

By substituting p = 50 cm and q = 21.3 cm, we find

h_i = - h_o \frac{21.3 cm}{50 cm}=-0.43 h_o

So we notice that:

|h_i| < |h_o| : this means that the image is smaller than the object

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so, the correct option is:

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