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Ad libitum [116K]
3 years ago
6

Batteries are usually identified by their voltage.

Physics
1 answer:
Fynjy0 [20]3 years ago
7 0

Answer:

Explanation:

Given parameters:

Electrical energy  = 18J

Quantity of charge Q = 12C

Unknown:

Voltage  = ?

Solution:

The expression of electrical potential energy is given as:

      Electrical potential energy  = \frac{1}{2} c v²

c is the quantity of charge

v is the voltage

           18  =  \frac{1}{2}  x 12 x v²  

           18 = 6v²  

              v² = 3  

               v = 1.7v

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4 years ago
The World-War II battleship U.S.S Massachusetts used 16-inch guns whose barrel lengths were 15 m long. The shells each of mass 1
Vaselesa [24]

Answer:

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

Explanation:

Let suppose that shells are not experiencing any effect from non-conservative forces (i.e. friction, air viscosity) and changes in gravitational potential energy are negligible. The explosive force experienced by the shell inside the barrel can be estimated by Work-Energy Theorem, represented by the following formula:

F\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^{2}) (1)

Where:

F - Explosive force, measured in newtons.

\Delta s - Barrel length, measured in meters.

m - Mass of the shell, measured in kilograms.

v_{o}, v_{f} - Initial and final speeds of the shell, measured in meters per second.

If we know that m = 1250\,kg, v_{o} = 0\,\frac{m}{s}, v_{f} = 750\,\frac{m}{s} and \Delta s = 15\,m, then the explosive force experienced by the shell inside the barrel is:

F = \frac{m\cdot (v_{f}^{2}-v_{o}^{2})}{2\cdot \Delta s}

F = \frac{(1250\,kg)\cdot \left[\left(750\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}\right]}{2\cdot (15\,m)}

F = 23437500\,N

The explosive force experienced by the shell inside the barrel is 23437500 newtons.

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3 years ago
Using the image provided, which moon phase is next?
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Answer:

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