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3 years ago

A square sheet of rubber has sides that are 20 cm long. What is the area of the square of rubber in cm squared?

Physics

1 answer:

Alinara [238K]3 years ago

6 0

Answer:

400cm^2

Explanation:

sides are 20cm long Area for a square is a squared

since all the lides are of equal length you can just choose one side.

20squared is 400

20 x 20 = 400cm squared

Hope this helps :)

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In a racquetball game, your opponent repeatedly hits the same type of serve off the front wall and it lands in approximately the

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2 years ago

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A boat takes 3.0 hours to travel 50 km down a river, then 5.4 hours to return. Determine the speed of the water in the river.

Nutka1998 [239]

Answer:

3.7 km/h

Explanation:

Let's call v the proper speed of the boat and v' the speed of the water in the river.

When the boat travels in the direction of the current, the speed of the boat is:

v + v'

And it covers 50 km in 3 h, so we can write

$v+v' = \frac{50 km}{3 h}=16.7 km/h$ (1)

When the boat travels in the opposite direction, the speed of the boat is

v - v'

And it covers 50 km in 5.4 h, so

$v-v'=\frac{50 km}{5.4 h}=9.3 km/h$ (2)

So we have a system of two equations: by solving them simultaneously, we find the value of v and v':

$v+v'=16.7 \\v-v'=9.3$

Subtracting the second equation from the first one we get:

$(v+v')-(v-v')=16.7-9.3\\2v'=7.4\\v'=3.7$

So, the speed of the water is 3.7 km/h.

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3 years ago

Consider the following reaction proceeding at 298.15 K: Cu(s)+2Ag+(aq,0.15 M)⟶Cu2+(aq, 1.14 M)+2Ag(s) If the standard reduction

lutik1710 [3]

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,

$Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)$

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

$E^o_{[Cu^{2+}/Cu]}=0.34V$

$E^o_{[Ag^{+}/Ag]}=0.80V$

$E^o=E^o_{[Ag^{+}/Ag]}-E^o_{[Cu^{2+}/Cu]}$

$E^o=0.80V-(0.34V)=0.46V$

Now we have to calculate the concentration of cell potential for this cell.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cu^{2+}][Ag]^2}{[Cu][Ag^+]^2}$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_{cell}$ = ?

Now put all the given values in the above equation, we get:

$E_{cell}=0.46-\frac{0.0592}{2}\log \frac{(1.14)\times (1)^2}{(1)\times (0.15)}$

$E_{cell}=0.434V$

Therefore, the cell potential for this cell 0.434 V

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