Ask question

Ask question

All categories

2 years ago

15

Can anyone give me the description of the element mercury is positioned on the periodic table of elements and a list of other el

ements that have similar properties. I need this ASAP. Thanks.

1 answer:

garri49 [273]2 years ago

3 0

I hope this helps you good luck

You might be interested in

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

It takes 200 N to move a box 10 meters in 8 seconds. How much power is

svetoff [14.1K]

Answer:

A

Explanation:

The equation of power is defined as Power = Workdone/Time Taken

And workdone = Force x Distance so using these equations we get they workdone is, 200x 10 = 2000Nm.

Dividing workdone with time will yield power, 2000 ÷ 8 = 250 Nm/s = 250W.

8 0

3 years ago

Given that two vectors A = 5i-7j-3k, B = -4i+4j-8k find A×B

WARRIOR [948]

$\textbf{A}×\textbf{B}= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

Explanation:

Given:

$\textbf{A} = 5\hat{\textbf{i}} - 7\hat{\textbf{j}} - 3\hat{\textbf{k}}$

$\textbf{B} = -4\hat{\textbf{i}} + 4\hat{\textbf{j}} - 8\hat{\textbf{k}}$

The cross product $\textbf{A}×\textbf{B}$ is given by

$\textbf{A}×\textbf{B} = \left|\begin{array}{ccc}\hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\\:\:5 & -7 & -3 \\ -4 & \:\:4 & -8 \\ \end{array}\right|$

$= \left|\begin{array}{cc}-7 & -3\\\:4 & -8\\ \end{array}\right|\:\hat{\textbf{i}}\:+\:\left|\begin{array}{cc}-3 & \:\:5\\-8 & -4\\ \end{array}\right|\:\hat{\textbf{j}}\:+\: \left|\begin{array}{cc}\:\:5 & -7\\-4 & \:\:4\\ \end{array}\right|\:\hat{\textbf{k}}$

$= 68\hat{\textbf{i}} + 52\hat{\textbf{j}} - 8\hat{\textbf{k}}$

5 0

2 years ago

A farsighted boy has a near point at 2.3 m and requires eyeglasses to correct his vision. Corrective lenses are available in inc

tino4ka555 [31]

Answer:

P = 3.5 D

Explanation:

As we know that convex lens is to be used to make the near point of eye to be correct

So we will have

$\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}$

here we have

$d_i = 2.3 m = 230 cm$

$d_o = 25 cm$

now plug in all values into the formula

$-\frac{1}{230} + \frac{1}{25} = \frac{1}{f}$

$f = 28 cm$

now for power of lens

$P = \frac{1}{f}$

$P = \frac{1}{0.28} = 3.5 D$

so the power in dioptre is

P = 3.5 D

5 0

2 years ago

Gas liquids will both expand to fll their containers true or false

MrMuchimi

It Is False. Liquid will not expand to fill it's container, while gas will.

6 0

3 years ago