The voltage across a group of cells connected in series is the sum of the voltages of each individual cell.
In the given paragraph, it states that when two or more cells are connected together side by side, the voltage across them is the sum of the voltage of each cell. This means that if you have two cells connected together, the total voltage across them would be the sum of the voltage of each individual cell.
Let's consider an example where you have two cells connected in series and the total voltage across them is 4.5 volts. In this case, it means that the voltage of each individual cell is not necessarily 4.5 volts. The total voltage is the combination of the voltages of both cells.
To find the voltage of each individual cell, you need to know the total voltage and the number of cells connected. In this example, if you have two cells and the total voltage is 4.5 volts, it would mean that each cell has a voltage of 2.25 volts. This is because the total voltage is divided equally among the cells when they are connected in series.
To summarize, the voltage across a group of cells connected in series is the sum of the voltages of each individual cell. The individual voltage of each cell can be found by dividing the total voltage by the number of cells connected.
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A 900 pF capacitor is charged by 100 V battery. (a) How much electrostatic energy is stored by the capacitor? The capacitor is disconnected from the battery and connected to another 900 pF capacitor. How much is the electrostatic energy stored in the system?
Answer:
The electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules
Explanation:
The formula to calculate the electrostatic energy stored in a capacitor is given by:
E = (1/2) * C * V^2
where:
E is the electrostatic energy,
C is the capacitance, and
V is the voltage across the capacitor.
(a) For the first scenario, where a 900 pF capacitor is charged by a 100 V battery:
C = 900 pF = 900 * 10^(-12) F
V = 100 V
Using the formula, we can calculate the electrostatic energy stored in the capacitor:
E = (1/2) * C * V^2
E = (1/2) * (900 * 10^(-12)) * (100^2)
E = 0.045 J
Therefore, the electrostatic energy stored by the capacitor in the first scenario is 0.045 Joules.
(b) In the second scenario, when the first capacitor is disconnected from the battery and connected to another 900 pF capacitor, the total capacitance in the system becomes:
C_total = C1 + C2
C_total = 900 pF + 900 pF
C_total = 1800 pF = 1800 * 10^(-12) F
The voltage across the capacitors remains the same, as they are connected in parallel.
Using the formula for electrostatic energy, we can calculate the new energy stored in the system:
E_total = (1/2) * C_total * V^2
E_total = (1/2) * (1800 * 10^(-12)) * (100^2)
E_total = 0.09 J
Therefore, the electrostatic energy stored in the system after connecting the two capacitors is 0.09 Joules.
A 40kg child is on a swing. At the bottom of a swing, the child attains a speed
of 4.0 meters per second. If the chain holding the swing is 4.0m long, what is
the Force of Contact by the chain on the child at the bottom of the swing
(assume the chain has no mass)?
The force of contact by the chain on the child at the bottom of the swing is 160 N. This is the minimum force required to keep the child moving in a circular path with a constant speed of 4.0 m/s. Any lesser force and the child will move away from the circular path and slow down.
At the bottom of the swing, the child has a velocity of 4.0 m/s and is being pulled downwards by the force of gravity. The tension in the chain holding the swing provides the necessary centripetal force to keep the child moving in a circular path. The centripetal force required is given by the formula F = (mv^2)/r, where F is the force, m is the mass of the child, v is the velocity, and r is the radius of the circular path. In this case, the radius is equal to the length of the chain, which is 4.0 m.
Substituting the given values, we get:
F = (40 kg x (4.0 m/s)^2)/4.0 m
F = 160 N
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The surface of the exit road is horizontal, not banked. (See figure.) If the static
friction between the tires and the surface of the road is us = 0.688 and the
maximum speed with which the car can exit the highway safely without sliding is
25.2 m/s, what is the radius of curvature of a highway exit, r?
The radius of curvature of the highway exit is approximately 220 km.
To find the radius of curvature of the highway exit, we can use the centripetal force equation:
F = mv^2 / r
where F is the maximum static friction force, m is the mass of the car, v is the maximum safe speed, and r is the radius of curvature.
We can solve for r by rearranging the equation:
r = mv^2 / F
Substituting the given values, we have:
r = (1000 kg)(25.2 m/s)^2 / (0.688)
r = 2.20 x 10^5 m
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(a) Define the following terms(1) internal energy (ii) an isovolumic process (b) state first law of thermodynamics (c
The definitions of Internal energy, Isovolumic process and the law of thermodynamics are:
(a) (i) Internal energy: Internal energy refers to the total energy stored within a system, including the kinetic and potential energies of its particles. It is a state function and depends only on the current state of the system, such as its temperature, pressure, and volume.
(ii) Isovolumic process: An isovolumic or isochoric process is a thermodynamic process where the volume of the system remains constant. During an isovolumic process, no work is done by or on the system through expansion or compression.
(b) The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed; it can only be transferred or converted from one form to another. The first law of thermodynamics is expressed mathematically as ΔU = Q - W, where ΔU is the change in internal energy of the system, Q is the heat transferred into or out of the system, and W is the work done on or by the system.
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