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Ede4ka [16]
2 years ago
14

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So t

he efficiency of the ramp would be ________________________ with the use of the dolly.
Physics
1 answer:
FromTheMoon [43]2 years ago
4 0

While pushing a refrigerator up a ramp, there would be less friction to overcome if you used a dolly because of the wheels. So the efficiency of the ramp would be greater with the use of the dolly.

<u>Explanation:</u>

The ramp provides for a wedge setup, so when the refrigerator is being pushed up the ramp, the friction on the ramp would equally distributed to the refrigerator when pushed wholly as the friction applied is a product of the coefficient of friction with the Normal reaction of the weight in contact with the surface.

Therefore, when the dolly is used, the surface in contact becomes divided and thus, the friction applied would become less due to though coefficient of friction remains the same, the normal reaction of the weight becomes small and so does the applied friction. Therefore, efficiency of the ramp increases as the friction is less and work done is more.

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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
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Answer : The cell potential for this cell 0.434 V

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The balanced cell reaction will be,  

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

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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}

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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)}

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3 years ago
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