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3 years ago

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What is the difference between the number of electrons in an atom of Iodine (I) and the number of electrons in an atom of Alumin

um (Al)?

1 answer:

o-na [289]3 years ago

5 0

Answer:

Iodine is element number 53. Iodine has 53 protons and 53 electrons (because it has no charge the number of protons equals the number of electrons). Iodine-131 has 131 protons and neutrons.

Explanation:

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4 years ago

20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to

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3 years ago

A tiny object carrying a charge of +44 μC and a second tiny charged object are initially very far apart. If it takes 21 J of wor

STatiana [176]

Answer:

The magnitude of the second charge is $\rm 1.062\times 10^{-7}\ C$ or $\rm 0.1062\ \mu C.$

Explanation:

The work done in bringing a charged particle from one point to another in the presence of some electric field is equal to the change in the electric potential energy of the charge in moving from one point to another.

The electric potential energy of some charge $q_o$ at a point in the electric field of another charge $q$ is given by the product of the amount of charge $q_o$ and electric potential at that point due to the charge $q$ .

$U = q_o\ V.$

The electric potential at that point is given by

$V = \dfrac{kq}{r}.$

where $k$ is the Coulomb's constant.

Therefore,

$U=q_o\ \dfrac{kq}{r}.$

Now, We have given two charges $q_1 = +44\ \mu C = +44\times 10^{-6}\ C$ and $q_2$ , whose value is to be found.

When the two charges are infinitely dar apart, the electric potential energy of the system is given by

$U_i = \dfrac{kq_1q_2}{\infty}=0.$

When the coordinates of position of the two charges are

$(x_1,\ y_1) = (1.00\ mm,\ 1.00\ mm).\\(x_2,\ y_2) = (1.00\ mm,\ 3.00\ mm).$

The distance between the two charges is given by

$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}=\sqrt{(1.00-1.00)^2+(3.00-1.00)^2}=2.00\ mm = 2.00\times 10^{-3}\ m.$

The electric potential energy of the charges in this configuration is given by

$U_f = \dfrac{kq_1q_2}{r}\\=\dfrac{(8.99\times 10^9)\times (+44\times 10^{-6})\times q_2}{2.00\times 10^{-3}}\\=1.9778\times 10^8\times q_2.$

The change in the electric potential energy of the system is equal to the work done to bring the system from inifinitely far apart position to given configuration.

Therefore,

$W = U_f-U_i\\21=(1.9778\times 10^8\times q_2)-0\\\Rightarrow q_2 = \dfrac{21}{1.9778\times 10^8}\\=1.062\times 10^{-7}\ C\\=0.1062\times 10^{-6}\ C\\=0.1062\ \mu C.$

6 0

3 years ago

Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and c

rodikova [14]

Answer:

Pressure of woman will be $99.87\times 10^5N/m^2$

Pressure of the elephant will be $1716560.50N/m^2$

Explanation:

We have given that mass of the woman m = 80 kg

Acceleration due to gravity $g=9.8m/sec^2$

Diameter of shoes = 1 cm =0.01 m

So radius $r=\frac{d}{2}=\frac{0.01}{2}=0.005m$

So area $A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2$

We know that force is given F = mg

So $F=80\times 9.8=784N$

Now we know that pressure is given by $P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2$

Now mass of elephant m = 5500 kg

So force of elephant = 5500×9.8 = 53900 N

Diameter = 20 cm

So radius r = 10 cm

So area will be $A=3.14\times 0.1^2=0.0314m^2$

So pressure will be $P=\frac{53900}{0.0314}=1716560.50N/m^2$

3 0

3 years ago