Answer:
9 meters
Explanation:
Given:
Mass of Avi is, 
Spring constant is, 
Compression in the spring is, 
Let the maximum height reached be 'h' m.
Now, as the spring is compressed, there is elastic potential energy stored in the spring. This elastic potential energy is transferred to Avi in the form of gravitational potential energy.
So, by law of conservation of energy, decrease in elastic potential energy is equal to increase in gravitational potential energy.
Decrease in elastic potential energy is given as:
Now, increase in gravitational potential energy is given as:
Now, increase in gravitational potential energy is equal decrease in elastic potential energy. Therefore,
Therefore, Avi will reach a maximum height of 9 meters.
1 volt = 1 joule per coulomb.
Current doesn't actually pass 'through' a battery.
But if it did, then each coulomb would gain or lose 6 joules in traversing 6 volts, depending on its sign, and whether it climbed or fell.
Answer:
0.281 m
Explanation:
Applying,
W = 0.5ke².................. Equation 1
Where W = Workdone by the stretched spring, k = spring constant, e = extension/ compression of the spring
make e the subject of the equation
e = √(2W/k)................ Equation 2
From the question,
Given: W = 1.240J, k = 31.50 N/m
Substitute these values into equation 2
e = √(2×1.240/31.50)
e = √(0.07873)
e = 0.281 m
The charge accumulated in 3.25 μF capacitor is 178.75 μC.
Answer:
Explanation:
In parallel connection, the voltage drop across any passive devices like capacitor or resistor will be constant. So the current flow will be varying in case of parallel connections of capacitors or resistors.
As the capacitance is the measure of amount of charge or current generated for a given amount of voltage, it is directly proportional to the charge or current and inversely proportional to voltage.
C = Q/V
Here the charge accumulated in a capacitor of capacitance 3.25 microfarad need to be determined which is in parallel connection with another capacitor. So the voltage through both the capacitor will be equal to the voltage of the battery which is stated as 55 V.
3.25×
= Q/55
Q = 3.25 * 55 = 178.75 μC
So the charge accumulated in 3.25 μF capacitor is 178.75 μC.
