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2 years ago

Which of the following is a chemical change?

Physics

2 answers:

e-lub [12.9K]2 years ago

4 0

A screw driver rusting when left in a pail of water is a chemical reaction

Kisachek [45]2 years ago

4 0

Answer:

a screwdriver rusting when left in a pail of water

Explanation:

In case of chemical change we can say that there must be a chemical reaction due to which there will be a change in the chemical reactants.

As we can see the given options

1). wind breaking down a rock into smaller pieces

There is no chemical reaction between them and it just a physical change of big size to smaller size

2). a leaf falling off of a tree during a storm

As we know that here also leaf detach from the tree so there is only physical change again.

3). a screwdriver rusting when left in a pail of water

Here we will have a chemical change due to which iron made its oxide due to chemical reaction of iron and oxygen

4). a block of ice melting in the sun

This is also physical change in which water molecules change from its solid state to liquid state

So here correct answer will be

a screwdriver rusting when left iwind breaking down a rock into smaller pieces

a leaf falling off of a tree during a storm

a screwdriver rusting when left in a pail of water

a block of ice melting in the sunwind breaking down a rock into smaller pieces

a leaf falling off of a tree during a storm

a screwdriver rusting when left in a pail of water

a block of ice melting in the sunwind breaking down a rock into smaller pieces

a leaf falling off of a tree during a storm

<em>a screwdriver rusting when left in a pail of water </em>

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3 years ago

A spherical capacitor contains a charge of 3.00 nC when connected to a potential difference of 230 V. If its plates are separate

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Answer:

Part(a): the capacitance is 0.013 nF.

Part(b): the radius of the inner sphere is 3.1 cm.

Part(c): the electric field just outside the surface of inner sphere is $\bf{2.81 \times 10^{4}~n~C^{-1}}$ .

Explanation:

We know that if 'a' and 'b' are the inner and outer radii of the shell respectively, 'Q' is the total charge contains by the capacitor subjected to a potential difference of 'V' and ' $\epsilon_{0}$ ' be the permittivity of free space, then the capacitance (C) of the spherical shell can be written as

$C = \dfrac{4 \pi \epsilon_{0}}{(\dfrac{1}{a} - \dfrac{1}{b})}~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Part(a):

Given, charge contained by the capacitor Q = 3.00 nC and potential to which it is subjected to is V = 230V.

So the capacitance (C) of the shell is

$C &=& \dfrac{Q}{V} = \dfrac{3 \times 10^{-90}~C}{230~V} = 1.3 \times 10^{-11}~F = 0.013~nF$

Part(b):

Given the inner radius of the outer shell b = 4.3 cm = 0.043 m. Therefore, from equation (1), rearranging the terms,

$&& \dfrac{1}{a} = \dfrac{1}{b} + \dfrac{1}{C/4 \pi \epsilon_{0}} = \dfrac{1}{0.043} + \dfrac{1}{1.3 \times 10^{-11} \times 9 \times 10^{9}} = 31.79\\&or,& a = \dfrac{1}{31.79}~m = 0.031~m = 3.1~cm$

Part(c):

If we apply Gauss' law of electrostatics, then

$&& E~4 \pi a^{2} = \dfrac{Q}{\epsilon_{0}}\\&or,& E = \dfrac{Q}{4 \pi \epsilon_{0}a^{2}}\\&or,& E = \dfrac{3 \times 10^{-9} \times 9 \times 10^{9}}{0.031^{2}}~N~C^{-1}\\&or,& E = 2.81 \times 10^{4}~N~C^{-1}$

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3 years ago