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blondinia [14]
3 years ago
15

list all the storage forms of energy that forms of energy that you are familiar with. for each storage form, give an example of

an object or situation that is storing this form of energy.
Physics
1 answer:
ella [17]3 years ago
8 0

Storage form of energy:

  • Potential energy
  • Nuclear energy
  • electrical energy
  • thermal energy
  • magnetic energy

Potential energy:

All stationary objects are having potential energy stored in it. This energy can be transferred in form of kinetic energy when it comes in the motion from rest. Example, An object placed at height h having potential energy in it. When it comes in motion from the rest the potential energy is converted into kinetic energy.

Nuclear energy:

Nuclear energy is energy that is stored in nucleus of any element. Example, fusion reaction on sun gives earth solar energy.

Electric energy:

Electrical energy is due to movement of the electrical charges. Example, In elctrical batteries electrical energy is stored.

Thermal energy:

Thermal energy is the internal energy of a substance that is transferred to other substance in the form of heat. Example, on heating water is a beaker stem energy is developed.

Magnetic energy:

Magnetic energy is the potential energy stored in the magnetic field. Example, using magnetic energy electric field is produced according to Faraday's law.


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Two traveling sinusoidal waves are described by the wave functions y1 = 4.85 sin [(4.35x − 1270t)] y2 = 4.85 sin [(4.35x − 1270t
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Answer:

Approximately 9.62.

Explanation:

y_1 = 4.85\, \sin[(4.35\, x - 1270\, t) + 0].

y_2 = 4.85\, \sin[(4.35\, x - 1270\, t) + (-0.250)].

Notice that sine waves y_1 and y_2 share the same frequency and wavelength. The only distinction between these two waves is the (-0.250) in y_2\!.

Therefore, the sum (y_1 + y_2) would still be a sine wave. The amplitude of (y_1 + y_2)\! could be found without using calculus.

Consider the sum-of-angle identity for sine:

\sin(a + b) = \sin(a) \cdot \cos(b) + \cos(a) \cdot \sin(b).

Compare the expression \sin(a + b) to y_2. Let a = (4.35\, x - 1270) and b = (-0.250). Apply the sum-of-angle identity of sine to rewrite y_2\!.

\begin{aligned}y_2 &= 4.85\, \sin[(\underbrace{4.35\, x - 1270\, t}_{a}) + (\underbrace{-0.250}_{b})]\\ &= 4.85 \, [\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Therefore, the sum (y_1 + y_2) would become:

\begin{aligned}& y_1 + y_2\\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t) \\ &\quad \quad \quad\;+\sin(4.35\, x - 1270\, t)\cdot \cos(-0.250) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \\[0.5em] &= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Consider: would it be possible to find m and c that satisfy the following hypothetical equation?

\begin{aligned}& (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c)\\&= 4.85\, [\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad\quad\; + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)] \end{aligned}.

Simplify this hypothetical equation:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\&=\sin(4.35\, x - 1270\, t)\cdot (1 + \cos(-0.250)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot \sin(-0.250)\end{aligned}.

Apply the sum-of-angle identity of sine to rewrite the left-hand side:

\begin{aligned}& m\cdot \sin((4.35\, x - 1270\, t) + c)\\[0.5em]&=m\, \sin(4.35\, x - 1270\, t)\cdot \cos(c) \\ &\quad\quad + m\, \cos(4.35\, x - 1270\, t)\cdot \sin(c) \\[0.5em] &=\sin(4.35\, x - 1270\, t)\cdot (m\, \cos(c)) \\ &\quad\quad + \cos(4.35\, x - 1270\, t)\cdot (m\, \sin(c)) \end{aligned}.

Compare this expression with the right-hand side. For this hypothetical equation to hold for all real x and t, the following should be satisfied:

\displaystyle 1 + \cos(-0.250) = m\, \cos(c), and

\displaystyle \sin(-0.250) = m\, \sin(c).

Consider the Pythagorean identity. For any real number a:

{\left(\sin(a)\right)}^{2} + {\left(\cos(a)\right)}^{2} = 1^2.

Make use of the Pythagorean identity to solve this system of equations for m. Square both sides of both equations:

\displaystyle 1 + 2\, \cos(-0.250) +  {\left(\cos(-0.250)\right)}^2= m^2\, {\left(\cos(c)\right)}^2.

\displaystyle {\left(\sin(-0.250)\right)}^{2} = m^2\, {\left(\sin(c)\right)}^2.

Take the sum of these two equations.

Left-hand side:

\begin{aligned}& 1 + 2\, \cos(-0.250) + \underbrace{{\left(\cos(-0.250)\right)}^2 + {\left(\sin(-0.250)\right)}^2}_{1}\\ &= 1 + 2\, \cos(-0.250) + 1 \\ &= 2 + 2\, \cos(-0.250) \end{aligned}.

Right-hand side:

\begin{aligned} &m^2\, {\left(\cos(c)\right)}^2 + m^2\, {\left(\sin(c)\right)}^2 \\ &= m^2\, \left( {\left(\sin(c)\right)}^2 +  {\left(\cos(c)\right)}^2\right)\\ &= m^2\end{aligned}.

Therefore:

m^2 = 2 + 2\, \cos(-0.250).

m = \sqrt{2 + 2\, \cos(-0.250)} \approx 1.98.

Substitute m = \sqrt{2 + 2\, \cos(-0.250)} back to the system to find c. However, notice that the exact value of c\! isn't required for finding the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c).

(Side note: one possible value of c is \displaystyle \arccos\left(\frac{1 + \cos(0.250)}{\sqrt{2 \times (1 + \cos(0.250))}}\right) \approx 0.125 radians.)

As long as \! c is a real number, the amplitude of (y_1 + y_2) = (4.85\, m)\cdot \sin((4.35\, x - 1270\, t) + c) would be equal to the absolute value of (4.85\, m).

Therefore, the amplitude of (y_1 + y_2) would be:

\begin{aligned}|4.85\, m| &= 4.85 \times \sqrt{2 + 2\, \cos(-0.250)} \\&\approx 9.62 \end{aligned}.

8 0
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How much work is done on a 75 newton bowling ball when you carry it horizontally across a 10 meter room
svlad2 [7]

F = force applied to hold the weight of the bowling ball = weight of the bowling ball = 75 N

d = distance through which the bowling ball is moved horizontally = 10 meter

θ = angle between the force in vertically upward direction  and displacement in horizontal direction = 90

W = work done on the bowling ball

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W =  (75) (10) (0)

W = 0 J

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