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2 years ago

According to the graph, how many atoms would remain after two half-lives?

Physics

1 answer:

Fudgin [204]2 years ago

4 0

Answer:

Let No be initial no of atoms

N = N0 / 2 after 1 half-life

N = N0 / 4 after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

7 0

3 years ago

What elements are good conductors of electricity?

kramer

In solids: All metals are good conductors of electricity as they contain free moving electrons. Non-metals doesn't conduct , but we consider Graphite the only non-metal that can conduct electricity for the presence of free moving electrons.

In Liquids ; Ionic compunds contains free moving ions , so they conduct electricity as well .

6 0

3 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns

Explanation:

We can solve this with the following relation:

$\frac{d}{x_{sunball-pinhole}}=\frac{D}{x_{sun-pinhole}}$

Where:

$d=17.91 mm =17.91(10)^{-3} m$ is the diameter of a dime

$D$ is the diameter of the Sun

$x_{sun-pinhole}=150,000,000 km=1.5(10)^{11} m$ is the distance between the Sun and the pinhole

$x_{sunball-pinhole}=100 d=1.791 m$ is the amount of dimes that fit in a distance between the sunball and the pinhole

Finding $D$ :

$D=\frac{d}{x_{sunball-pinhole}}x_{sun-pinhole}$

$D=\frac{17.91(10)^{-3} m}{1.791 m} 1.5(10)^{11} m$

$D=1.5(10)^{9} m$ This is roughly the diameter of the Sun

Now, the distance between the Earth and the Sun is one astronomical unit (1 AU), which is equal to:

$1 AU=149,597,870,700 m$

So, we have to divide this distance between $D$ in order to find how many suns could it fit in this distance:

$\frac{149,597,870,700 m}{1.5(10)^{9} m}=99.73 suns \approx 100 suns$

8 0

3 years ago

A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the

Goshia [24]

To solve this problem we will use the concepts related to power, defined as the amount of energy applied over a period of time.

The energy in this case is the accumulated in the form of potential energy, over a period of time. Thus we will have that the mathematical expression of the power can be expressed as

$P = \frac{E}{t}$

Here,

E = Energy

t = time

As the energy is equal to the potential Energy we have tat

$P = \frac{mgh}{t}$

The weight (mg) of the man is 700N, the height (h) is 8m and the time is 10s, then:

$P = \frac{700*8}{10}$

$P = 560W$

Therefore the correct answer is A.

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3 years ago

Which property of table salt is also a property of other ionic compounds

Jet001 [13]

s alluded to in the other answers, salt refers to any ionic compound that doesn't have “oxides” in it. Table salt is sodium chloride. Going down the periodic table, the first column contains lithium, sodium, potassium, rubidium, cesium, and francium. This group (alkali metals) of atoms (and their corresponding positive ions) gets larger in the order shown above. Therefore, their ionic bonds with chloride (or any nonmetal) gets smaller. The trend of their corresponding compounds is a decreasing hardness, decreasing melting point, decreasing boiling point, and decreasing thermal stability. These are the major periodic trends of these corresponding compounds. Other metal ions generally have higher positive charges on them. This makes the ionic bonds considerably larger and you can probably surmise most of their corresponding properties listed above. However, the details of their lattice structures may cause the overall trend to vary.

3 0

3 years ago

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