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3 years ago

The Curiosity rover now on Mars analyzed rocks and found magnesium to have the following isotopic composition.

Physics

1 answer:

emmasim [6.3K]3 years ago

7 0

Answer:

A. number of neutrons of Magnesium Mg = 13

B. The average mass of Mg = 22.29 amu

C. the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= $\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}$

= $\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}$

= $\mathtt{\dfrac{2229.17774 }{100}}$

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

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A particle undergoes two displacements. The first has a magnitude of 11 m and makes an angle of 82 ◦ with the positive x axis. T

Luden [163]

Answer:

$\theta = 211.7 degree$

Explanation:

First displacement of the particle is given as

$r_1$ = 11 m at 82 degree with positive X axis

so we can say

$\vec r_1 = 11 cos82 \hat i + 11 sin82 \hat j$

$\vec r_1 = 1.53\hat i + 10.9 \hat j$

resultant displacement of the particle after second displacement is given as

r = 8.7 m at 135 degree with positive X axis

so we can say

$r = 8.7 cos135\hat i + 8.7 sin135\hat j$

$r = -6.15 \hat i + 6.15 \hat j$

now we know that

$r = r_1 + r_2$

now we have

$r_2 = r - r_1$

so we will have

$r_2 = (-6.15 \hat i + 6.15 \hat j) - (1.53\hat i + 10.9 \hat j)$

$r_2 = -7.68 \hat i - 4.75 \hat j$

so angle of the second displacement is given as

$tan\theta = \frac{r_y}{r_x}$

$tan\theta = \frac{-4.75}{-7.68}$

$\theta = 211.7 degree$

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3 years ago

A 15.7 kg block is dragged over a rough, horizontal surface by a constant force of 83.1 N

Kay [80]

Answer:

The work done by the 83.1 N force is 4687.5 J.

The magnitude of the work done by the force of friction is 1187.5 J.

Explanation:

Given:

Mass of the block, $m=15.7$ kg

Force acting on it, $F=83.1$ N

Angle of application of force, $\theta = 25.7$ °

Displacement of the block, $d=62.6$ m

Coefficient of friction, $\mu =0.161$

Acceleration due to gravity, $g=9.8$ m/s²

Work done by a force is given as:

$Work,W=F\times d\times \cos\theta$

So, work done by the constant force is given as:

$W_{force}=83.1\times 62.6\times \cos(25.7)\\W_{force}=4687.5\textrm{ J}$

Now, in order to find work done by friction, we need to evaluate friction.

For the vertical direction, the net force is zero as there is no vertical motion.

Therefore,

$N + F\sin\theta = mg\\ N = mg-F\sin\theta$

Frictional force is given as:

$f=\mu N=\mu (mg-F \sin \theta)=0.161\times ((15.7\times 9.8)-(83.1\times \sin(25.7))=18.97\textrm{ N}$

Now, friction acts in the direction opposite to the displacement. Thus, angle between frictional force and displacement is 180°.

Therefore, work done by friction is:

$W_{fric}=f\times d\times \cos\theta_f\\W_{fric}=12.72\times 62.6\times \cos(-180)\\W_{fric}=18.97\times 62.6\times -1\\W_{fric}=-1187.5\textrm{ J}$

Negative sign indicates that frictional force is acting opposite to motion.

So, the magnitude of the work done by the force of friction is 1187.5 J.

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