A car is moving with a constant acceleration. At t = 5.0 s its velocity is 8.0 m/s and at t = 8.0 s its

ozzi

Detailed Explanation:

1) Rusting of Iron

4Fe + 3O2 + 2H2O -> 2Fe2O32H2O

Reactants :-
Fe = 4
O = 3 * 2 + 2 = 8
H = 2 * 2 = 4

Products :-
Fe = 2 * 2 = 4
O = 2 * 3 + 2 = 8
H = 2 * 2 = 4

2) Fermentation of sucrose…

C12H22O11 + H2O -> 4C2H5OH + 4CO2

Reactants :-
C = 12
H = 22 + 2 = 24
O = 11 + 1 = 12

Products :-
C = 4 * 2 + 4 = 12
H = 4 * 5 + 4 = 24
O = 4 * 2 + 4 = 12

Looking closely at the way I have taken the total number of elements on the reactants and products side, you can solve the rest.

All the Best!

8 0

2 years ago

Read 2 more answers

Bits of paper are attracted to an electrified comb or rod, even though they have nonet charge. How is this possible?

SashulF [63]

Explanation:

An electrified comb is charged comb ( let say by running it through the hair) and when it is brought in the proximity of pieces of paper, the pieces tend to cling to it. This happens because the charged comb induces an opposite charge in the paper pieces and as opposite charges attract each other, the pieces are clinged.

7 0

3 years ago

How are the different firework shapes produced

AleksAgata [21]

To create the shapes, stars are arranged on a piece of cardboard in the desired configuration. If the stars are placed in a smiley face pattern on the cardboard, for example, they will explode into a smiley face in the sky. In fact, you may see several smiley faces in the sky at one time.

Your Welcome

8 0

2 years ago

A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne

Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

$F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz$

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0

3 years ago

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then:

$I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}$

$I = 1370.05A$

The current is too high to be transported which would make the case not feasible.

8 0

2 years ago