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1 answer:
Answer:
Total resistance is 19.6 Ω for the circuit .
Explanation:
Given:
Six resistors of a circuit namely to
.
Where
and
are in series.
and
are in parallel.
and
are in parallel.
We know that continuous resistor which are in series are added up to find the equivalent resistance.
Similarly resistors which are arranged in parallel their equivalent resistance is for the case of
and
.
According to the question:
Total resistance = R(equivalent) :
⇒
⇒
⇒
⇒
⇒ Ω
So the total resistance of the circuit depicted is 19.58 Ω approximated to nearest tenth that is 19.6 Ω
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There are two things that you should remember while dealing with the "Lever Mechanical Advantage" problems:
1) The Effort Arm;
2) The Resistance Arm.
Some books label the Effort Arm as in-lever arm and the Resistance Arm as out-lever arm. (Physics Jargon that you need to remember in order to solve problems)
The Effort Arm is that "part" of the lever where the force can be applied. The Resistance Arm is where some mass is placed. In the diagram, as you can see, the mass is placed on one arm of the lever. Therefore, it is the Resistance Arm.
Now the formula for the "Mechanical Advantage(MA)" is:
Where
is the length of the Effort Arm(the subscript "e" stands for Effort), and 
stands for the length of the Resistance Arm(here "r" stands for Resistance).
Plug in the values:
= 15m.
= 7m.
Therefore, / = 15/7 = 2.143 = MA
The correct answer is option C(2.14).
-i
1) The Effort Arm;
2) The Resistance Arm.
Some books label the Effort Arm as in-lever arm and the Resistance Arm as out-lever arm. (Physics Jargon that you need to remember in order to solve problems)
The Effort Arm is that "part" of the lever where the force can be applied. The Resistance Arm is where some mass is placed. In the diagram, as you can see, the mass is placed on one arm of the lever. Therefore, it is the Resistance Arm.
Now the formula for the "Mechanical Advantage(MA)" is:
Where
Plug in the values:
Therefore,
The correct answer is option C(2.14).
-i
Answer:
F_ab = 260.17 N
Explanation:
Given:
- Moment of force F about CD, (M)_cd = 57 Nm
Find:
- First we will write down the position vectors of points A, B , C , D:
- We will take left and bottom most corner of cube to be the origin.
- The unit vectors i , j , k are along vertical planes and outside the plane, respectively.
- The position vectors wrt to the origin are:
Point A = 0.2 k
Point B = 0.4 i + 0.2 j
Point C = 0.2 j + 0.4 k
Point D = 0.4 i + 0.4 k
- Now we will determine the Force vector F_ab along vector AB.
vec (AB) = B - A = 0.4 i + 0.2 j - 0.2 k
unit (AB) = 0.4 i + 0.2 j - 0.2 k / sqrt ( 0.4^2 + 2*0.2^2)
= [5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
Hence,
vec(F_ab) = Fab*[5 / sqrt(6)] * ( 0.4 i + 0.2 j - 0.2 k )
- Now, form a unit vector along the line CD:
vec(CD) = D - C = 0.4 i - 0.2 j
unit (CD) = 0.4 i - 0.2 j / sqrt ( 0.4^2 + 0.2^2)
= [sqrt(5)]*(0.4 i - 0.2 j)
- Now select a point on line CD, lets say C. Find the moment arm from line of action of force along AB and line CD. Hence, vector AC is:
vec(AC) =r_ac = C - A = 0.2 j + 0.2 k
- Now the moment about a line CD due to force is:
(M)_cd = unit(CD) . ( r_ac x vec(F_ab) )
The cross product of r_ac and vec(F_ab) is as follows:
(M)_c = ( r_ac x vec(F_ab) ) :
(M)_c = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k]
The dot product of (M)_c and unit (CD) is as follows:
(M)_cd = unit(CD) . (M)_c :
(M)_cd = F_ab[- sqrt(6)/15 i + sqrt(6)/15 j - sqrt(6)/15 k] . [sqrt(5)]*(0.4 i - 0.2 j)
(M)_cd = F_ab*(sqrt(30) / 25)
- The given magnitude of the moment is (M)_cd. Calculate F_ab:
57 = F_ab*(sqrt(30) / 25)
F_ab = 260.17 N
Answer:
Temperature of the hot reservoir is 1540K
Explanation:
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K