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3 years ago

A list of five compounds potassium can form

1 answer:

5 0

Potassium can form so many compounds.
I will write important ones which you will surely come across in future only if you take Chemistry

1. KMnO4

It is a widely used oxidising titrant. It acts as self indicator for end point detection and is a very strong oxidant.

2. KNO3

In galvanic cell, salt bridge(a U shaped glass tube filled with a concentrated solution of electrolyte usually KNO3) is used a bridge between two concentrated solutions.

3. KCl

Again same as 2 ^.^

4. KOH

It is alkaline in nature and can be easily formed in laboratory

5. K2S2O8

Alkaline potassium persulphate (K2S2O8) is used to oxidise monohydric phenols to dihydric phenols.

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A student wearing frictionless in-line skates on a horizontal surface is pushed, starting from rest, by a friend with a constant

grin007 [14]

Answer:

Physics

Explanation:

Explanation:

We can use the Theorem of Work (W) and Kinetic Energy (K):

W=ΔK=Kf−Ki

it basically tells us that the work done on our system will show up as change in Kinetic Energy:

We know that the initial Kinetic Energy, Ki=12mv2i, is zero (starting from rest) while the final will be equal to 352J; Work will be force time displacement. so we get:

F⋅d=Ff

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d=35245=7.8≈8m

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2 years ago

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What is the amount of displacement of a runner who runs exactly 2 laps around a 400 meter track?

Alex787 [66]

Displacement is the distance and direction from the start point to the end point. Our runner finished exactly where he started. His displacement is zero.

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3 years ago

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At what height h above the ground does the projectile have a speed of 0.5v?

maw [93]

Answer:

$h=\dfrac{3v^2}{8g}$

Explanation:

It is given that,

Speed of the projectile is 0.5 v. Let h is the height above the ground. Using the first equation of motion to find it.

$v=u+at$

$v=u-gt$

Initial speed of the projectile is v and final speed is 0.5 v.

$0.5v=v-gt$

$t=\dfrac{v}{2g}$

g is the acceleration due to gravity

Let h is the height above the ground. Using the second equation of motion as :

$h=vt-\dfrac{1}{2}gt^2$

$h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2$

$h=\dfrac{3v^2}{8g}$

So, the height of the projectile above the ground is $\dfrac{3v^2}{8g}$ . Hence, this is the required solution.

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3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what

pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector $F_{net}$ = vector Fe + vector Fn,