Can somebody help me on this mcq question?

What property do the following elements have in common? Li, C, and F A) They are poor conductors of electricity. B) Each element

Aloiza [94]

Answer: Option (D) is the correct answer.

Explanation:

The given elements Li, C and F are all second period elements. So, when we move from left to right across a period then there occurs increase in number of valence electrons as there occurs increase in total number of electrons.

So, it means more electrons are added to the same energy level.

Thus, we can conclude that a property of valence electrons for each element is located in the same energy level is common in the given elements.

7 0

3 years ago

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

A defibrillator is used during a heart attack to restore the heart to its normal beating pattern. A defibrillator passes 18 A of

miskamm [114]

Answer:

q = 0.036 C

Explanation:

Given that,

Current passes through a defibrillator, I = 18 A

Time, t = 2 ms

We need to find the charge moved during this time. We know that,

Electric current = charge/time

$q=It$

Put all the values,

$q=18\times 0.002\\\\q=0.036\ C$

So, 0.036 C of charge moves during this time.

8 0

2 years ago

Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.

Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

$L = L_0 \sqrt{1-\frac{v^2}{c^2} }$

L = 1 light year

L₀ = 100,000 light year

$1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }$

$1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }$

$\frac{1}{100,000} = \sqrt{1-\frac{v^2}{c^2} }$

$v = 0.999999995$ $c$

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

#SPJ4

5 0

1 year ago

What is hooke's law? does it apply to elastic materials or to inelastic materials?

zysi [14]

When you talk about Hooke's law, it always have to do something with springs. Hooke's Law, from Robert Hooke, saw a relation between the force applied to the spring and the extension of its length. The equation is: F = kx, where k is the spring constant and x is the displacement of the original and stretched lengths. In other words, x is the length of deformation. Hence, the object must be elastic to come up with a displacement or deformation, in the first place. Then, the Hooke's Law is only applicable to elastic materials.

6 0

3 years ago