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3 years ago

Explain why the atomic mass of an element is a weighted-average mass

1 answer:

3 0

The isotopes contribute to the average atomic mass based on their abundance. The result is that the "average" mass for the atoms of an element is dictated by the most abundant or common isotope. The average atomic mass for carbon is 12.0107 amu. The atomic mass as displayed on the periodic table is a weighted average relative atomic mass of the naturally occuring isotopes of that element. An isotope is an element with the same number of protons but a different number of neutrons For example - Carbon naturally occurs in isotopes C12, C13 and C14 with abundances of 98.9% 1.1% and 'trace' respectively. the average mass is then calculated by 12*98.9%+13*1.1% = 12.01g/mol

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Facts about conolizing into mars?

Stolb23 [73]

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Permanent human habitation on a planetary body other than the Earth is one of science fiction's most prevalent themes. As technology has advanced, and concerns about the future of humanity on Earth have increased, the argument that space colonization is an achievable and worthwhile goal has gained momentum. Other reasons for colonizing space include economic interests, long-term scientific research best carried out by humans as opposed to robotic probes, and sheer curiosity.

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Many organizations support the colonization of Mars. They have also given different reasons and ways humans can live on Mars. One of the oldest organizations is the Mars Society. They promote a NASA program that supports human colonies on Mars. The Mars Society have set up Mars analog research stations in Canada and the United States. All other organizations include MarsDrive, who wants to help fund settlements on Mars, and Mars to Stay. Mars to Stay advocates settlements on Mars. In June 2012, Mars One released a statement that they believe could help start a colony on Mars by 2023.

3 0

2 years ago

A 10 N board of uniform density is 5 meters long. It is supported on the left by a string bearing a 3 N upward force. In order t

NikAS [45]

Answer:

C. $\frac{25}{7}$ m

Explanation:

We are given that

Weight of board=w=10 N

Length of board=L=5 m

Tension in the string=T=3 N

Applied upward force=F=7 N

We have to find the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

Let r be the distance at which its left wedge would they need to place this force in order for the board to be in static equilibrium.

The board is uniform therefore, the center of board is the mid- point of board.

Therefore, the lever arm of weight= $r_1=\frac{L}{2}=\frac{5}{2}m$

Now, the torque exerted by the weight of the board

$\tau_1=Force\times perpendicular\;distance=10\times \frac{5}{2}=25 N$

The torque exerted by applied force= $\tau_2=7\times r=7r$

In static equilibrium

The sum of rotational forces=0

$\tau_1+\tau_2=0$

The two rotational force act in opposite direction therefore,

$\tau_2=-7r$

Substitute the values

$25-7r=0$

$7r=25$

$r=\frac{25}{7}$ m

Hence, option C is true.

7 0

3 years ago

A proton is accelerated to 225V. Its de-Broglie wavelength is:

marin [14]

Answer:

The value is $\lambda = 1.9109 *10^{-12} \ m$

Explanation:

From the question we are told that

The potential of the proton is $V = 225 \ V$

Generally the momentum of the particle is mathematically represented as

$p = \sqrt{ 2 * m * V * e }$

Here e is the charge on the proton with value

$e = 1.60 *10^{-19} \ C$

m is the mass of the proton with value $m = 1.67 *10^{-27} \ kg$

$p = \sqrt{ 2 * (1.67*10^{-27} ) * 225 * 1.6*10^{-19}}$

=> $p = 3.4676 *10^{-22} \ kg \cdot m/ s$

So the de-Broglie wavelength isis mathematically represented as

$\lambda = \frac{h}{p}$

Here h is the Planck's constant with value

$h = 6.626 *10^{-34} \ J\cdot s$

=> $\lambda = \frac{6.626 *10^{-34}}{3.4676 *10^{-22} }$

=> $\lambda = 1.9109 *10^{-12} \ m$

6 0

3 years ago

Compare the KE of an 80 kg person walking at 4 m/s with that of the same person moving at 6 m/s.

Alla [95]

The correct answer is:
<span>KE at 4 m/s less than KE at 6 m/s

In fact, the formula for the kinetic energy is:
</span> $K= \frac{1}{2} mv^2$
<span>where m is the mass of the person and v his speed.

When the person is moving at 4 m/s, his kinetic energy is:
</span> $K= \frac{1}{2} (80 kg)(4 m/s)^2=640 J$
instead, when he's moving at 6 m/s, his kinetic energy is:
$K= \frac{1}{2}(80 kg)(6 m/s)^2=1440 J$

So, the kinetic energy at 4 m/s is less than the kinetic energy at 6 m/s.

7 0

3 years ago

Need help with these 4 questions

dimaraw [331]

Answer: 1. B

2.A

3.B

4. A

Explanation:

4 0

3 years ago