We will use two definitions to solve this problem. The first will be given by the conservation of energy, whereby the change in kinetic energy must be equivalent to work. At the same time, work can be defined as the product between the force by the distance traveled. By matching these two expressions and clearing for the Force we can find the desired variable.
Thus the force acting on the sled is,
Replacing,
Therefore the Force acting on the sled is 32N
Okay so here's the approach I took:
The potential difference in each of the circuits must be the same so if we derive equations for both the potential differences we can set them equal to each other and solve for R1:
In the first circuit
V = 2.2(R1)
In the second we have to find the equivalent resistor, since they are connected in series:
1/R1 + 1/R2 + 1/R3... = Rt
We have R2 so...
1/R1 + 1/3.1 = Rt
1/R1 + 0.323 = Rt
So...
V = 1.4(1/R1 + 0.323)
Set those equal:
2.2R1 = 1.4(1/R1 + 0.323)
2.2R1 = 1.4(1/R1) + 0.4522
Now multiply everything by R1 so we can combine like terms:
2.2R1^2 = 1.4 + 0.4522R1
Isolate to form a quadratic
2.2R1^2 - 0.4522R1 - 1.4 = 0
Solving this quadratic:
R1 = 0.90708 or R1 = -0.701
Since R cannot be negative
R1 = 0.907 ohms
Answer:
Explanation:
Given that,
We have two capacitors connected in series
C1=2.0-μF
C2=4.0-μF
Then the equivalent of their series connection
1/Ceq = ½ + ¼
1/Ceq= (2+1)/4
1/Ceq=¾
Taking the reciprocal
Ceq= 4/3 μF
The capacitors are connected to a battery of 1kv
V=1000Volts
We know that,
Q=CV
Where Q is charge
C is capacitance and
V is voltage
Then, Q=4/3 ×1000
Q=4000/3 -μC
Since the capacitors are in series, then the charge pass through them, so each charge on the capacitors are 4000/3 μF
After the capacitor has been charge, the capacitor are disconnect and reconnected in parallel to each other,
For parallel connection, they have the same voltage but different charges.
When connected in parallel, there is a charge redistribution,
And the total charge will be 2•4000/3=8000/3 -μF
Then, Q1 +Q2= 8000/3 μF
Now the charge on each capacitor will be, let them have a common voltage V
Q=CV
Then, Q1=C1V
Q1= 2×V=2V
Q2= 4×V=4V
Then, Q1+Q2=8000/3
4V+2V=8000/3
6V=8000/3
V=8000/(3×6)
V=4000/9
V=444.44Volts
Now, Q1=2V
Q1=2×4000/9
Q1=8000/9 μF
Also, Q2=4V
Q2=4×4000/9
Q2=16000/9 μF
Answer:
c. As the thread is stretched, the coating thins and its resistance increases; as the thread is relaxed, the coating returns nearly to its original state.
Explanation:
This is in line with the result as shown in the data.
Stretching causes the coating to thin which in turn increases the resistance of the thread (conduction drops) releasing it backs returns the thread to nearly its initial values, but due to the strain already in the thread it does not return to its original values.
Answer:
Volume will increase by factor 2
So option (A) will be correct answer
Explanation:
Let initially the volume is V pressure is P and temperature is T
According to ideal gas equation , here n is number of moles and R is gas constant
So ....................eqn 1
Now pressure is doubled and temperature is quadrupled
So new volume ........eqn 2
Now comparing eqn 1 nad eqn 2
So volume will increase by factor 2
So option (A) will be correct answer