They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.
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Answer:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. The total momentum of the trolleys after separation is zero
c. The momentum of the 2 kg trolley after separation is 12 kg·m/s
d. The momentum of the 3 kg trolley is -12 kg·m/s
e. The velocity of the 3 kg trolley = -4 m/s
Explanation:
a. The total momentum of the trolleys which are at rest before the separation is zero
b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0
c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s
d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s
e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley
∴ The momentum of the 3 kg trolley = 3 kg × Velocity of the 3 kg trolley = -12 kg·m/s
The velocity of the 3 kg trolley = -12 kg·m/s/(3 kg) = -4 m/s
Answer:
Explanation:
a ) Earlier emf of cell applied on R₁ but now emf will be distributed among R₁ and R₂
Potential difference on R₁ will become less .
b ) Current is inversely proportional to resistance of the circuit. As resistance increases , current will be less . So current through R₁ will become less.
c )
When resistance is added in series , they are added up to obtain equivalent resistance . So equivalent resistance R₁₂ will be more than R₁ OR R₂.
Answer:
Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.
![-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-](https://tex.z-dn.net/?f=-1.5%5C%3B%5Cmu%5Ctext%7BF%7D-%5B%5Cbegin%7Barray%7D%7Bc%7D-%7B%5Cbf%202.0%5C%3B%5Cmu%5Ctext%7BF%7D%7D-%5C%5C-3.0%5C%3B%5Cmu%5Ctext%7BF%7D-%5Cend%7Barray%7D%5D-)
.
Explanation:
Consider four possible cases.
<h3>Case A: 12.0 V.</h3>
In case all three capacitors are connected in parallel, the 
capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.
<h3>Case B: 5.54 V.</h3>
![-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-](https://tex.z-dn.net/?f=-3.0%5C%3B%5Cmu%5Ctext%7BF%7D-%5B%5Cbegin%7Barray%7D%7Bc%7D-%7B%5Cbf%202.0%5C%3B%5Cmu%5Ctext%7BF%7D%7D-%5C%5C-1.5%5C%3B%5Cmu%5Ctext%7BF%7D-%5Cend%7Barray%7D%5D-)
In case the capacitor is connected in parallel with the 
capacitor, and the two capacitors in parallel is connected to the 
capacitor in series.
The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.
The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,
.
What will be the voltage across the 2.0 μF capacitor?
The charge stored in two capacitors in series is the same as the charge in each capacitor.
.
Voltage is the same across two capacitors in parallel.As a result,
.
<h3>Case C: 2.76 V.</h3>
.
Similarly,
- the effective capacitance of the two capacitors in parallel is 5.0 μF;
- the effective capacitance of the three capacitors, combined:
.
Charge stored:
.
Voltage:
.
<h3 /><h3>Case D: 4.00 V</h3>
.
Connect all three capacitors in series.
.
For each of the three capacitors:
.
For the capacitor:
.
Answer:(a) 50 N
(b)38.34 N
Explanation:
Given
Maximum tension(T) in line 50 N
(a)If line is moving up with constant velocity i.e. there is no acceleration
This will happen when Tension is equal to weight of Fish
T-mg=0
T=mg
Maximum weight in this case will be 50 N
(b)acceleration of magnitude 
T-mg=ma
m=3.91
Therefore weight is 
