Two forces of 83 pounds and 56 pounds act simultaneously on an object. The resultant forms an angle of 52° with the 56-pound for
ce. Find the angle formed between the two original forces, to the nearest 10th of a degree.
1 answer:
Answer:
95.9°
Explanation:
The diagram illustrating the action of the two forces on the object is given in the attached photo.
Using sine rule a/SineA = b/SineB, we can obtain the value of B° as shown in the attached photo as follow:
a/SineA = b/SineB,
83/Sine52 = 56/SineB
Cross multiply to express in linear form
83 x SineB = 56 x Sine52
Divide both side by 83
SineB = (56 x Sine52)/83
SineB = 0.5317
B = Sine^-1(0.5317)
B = 32.1°
Now, we can obtain the angle θ, between the two forces as shown in the attached photo as follow:
52° + B° + θ = 180° ( sum of angles in a triangle)
52° + 32.1° + θ = 180°
Collect like terms
θ = 180° - 52° - 32.1°
θ = 95.9°
Therefore, the angle between the two forces is 95.9°
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Missing data in the text of the exercise: The molar concentration of Zinc is 10 times the molar concentration of copper.
Solution:
1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
where
is the standard potential at the cathode, while 
is the standard potential at the anode. For a Daniel Cell, at the cathode we have copper: 
, while at the anode we have zinc: 
. Therefore, at standard temperature the electrode potential difference of the Daniel Cell is
2) To calculate
at any temperature T, we should use Nerst equation:
![E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}](https://tex.z-dn.net/?f=E%5E0%28T%29%3DE%5E0-%20%5Cfrac%7BR%20T%7D%7Bz%20F%7D%20%5Cln%20%20%5Cfrac%7B%5BZn%5D%7D%7B%5BCu%5D%7D%20%20)
where
is the temperature in our problem
is the number of electrons transferred in the cell's reaction
is the Faraday's constant
![[Zn]](https://tex.z-dn.net/?f=%5BZn%5D)
and ![[Cu]](https://tex.z-dn.net/?f=%5BCu%5D)
are the molar concentrations of zinc and in copper, and in our problem we have ![[Zn]=10[Cu]](https://tex.z-dn.net/?f=%5BZn%5D%3D10%5BCu%5D)
.
Using all these data inside the equation, and using
, in the end we find:
![E^0(T)=E^0- \frac{R T}{z F} \ln \frac{[Zn]}{[Cu]}=+1.053 V](https://tex.z-dn.net/?f=E%5E0%28T%29%3DE%5E0-%20%5Cfrac%7BR%20T%7D%7Bz%20F%7D%20%5Cln%20%5Cfrac%7B%5BZn%5D%7D%7B%5BCu%5D%7D%3D%2B1.053%20V)
Solution:
1) First of all, let's calculate the standard electrode potential difference at standard temperature. This is given by:
where
2) To calculate
where
Using all these data inside the equation, and using
Answer:
Plss see attached file
Explanation:
