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3 years ago

PLLLLLLLLLLLZ HELP

Physics

1 answer:

Katyanochek1 [597]3 years ago

6 0

The second choice is the correct one.

You can spot it just from the units. An equation can't be true unless it has the same units on both sides.

Wavelength has units of (meters).

Frequency has units of (per second).

When you multiply them, you get (meters per second), the unit of speed, and the 'c' on the other side is speed.

None of the other choices has units of speed on the right side.

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food product with 10 kg mass is being transported to thesurface of the moon,where the acceleration due to gravity is1.624 m/s2;

larisa [96]

Answer:

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

Explanation:

g = Acceleration due to gravity

m = Mass = 10 kg

Weight on Earth

$W=mg\\\Rightarrow W=10\times 9.81\\\Rightarrow W=98.1\ N$

Converting to lbf

$98.1\times 0.22481=22.053861\ lbf$

On Moon

$W=10\times 1.624\\\Rightarrow W=16.24\ N$

Converting to lbf

$16.24\times 0.22481=3.6509144\ lbf$

In SI units 98.1 N, 16.24 N

English units 22.053861 lbf, 3.6509144 lbf

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2 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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