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gregori [183]
3 years ago
8

If the total energy of a system is 2000J, and the kinetic energy is 500J, what does the potential energy in the system have to b

e?
Physics
1 answer:
Lyrx [107]3 years ago
7 0

Answer:

Potential energy in the system = 1,500 J

Explanation:

Given:

Total energy of a system = 2,000 J

Kinetic energy = 500 J

Find:

Potential energy in the system

Computation:

Total energy of a system = Kinetic energy + Potential energy

2,000 J = 500 J + Potential energy in the system

Potential energy in the system = 1,500 J

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3 years ago
A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?
serious [3.7K]

Answer :  The wavelength of photon is, 4.63\times 10^{2}nm

Explanation : Given,

Energy of photon = 4.29\times 10^{-19}J

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm

Conversion used : 1nm=10^{-9}m

Therefore, the wavelength of photon is, 4.63\times 10^{2}nm

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3 years ago
All of the following elements produce magnetic fields except
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The answer should be copper. hope this helps :D
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3 years ago
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The train passes point A with a speed of 30 m/s and begins to decrease its speed at a constant rate of at = - 0.25 m/s^2. Determ
prisoha [69]

Explanation:

At point B, the velocity speed of the train is as follows.

          \nu^{2}_{B} = \nu^{2}_{A} + 2a_{t} (s_{B} - s_{A})

                           = (30)^{2} + 2(-0.25(412 - 0))

                           = 26.34 m/s

Now, we will calculate the first derivative of the equation of train.

          y = 200 e^{\frac{x}{1000}}

      \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}

Now, second derivative of the train is calculated as follows.

         \frac{dy}{dx} = 0.2 e^{\frac{x}{1000}}      

       \frac{d^{2}y}{dx^{2}} = 0.2 (10^{-3}) e^{\frac{x}{1000}}    

Radius of curvature of the train is as follows.  

   \rho = \frac{[1 + (\frac{dy}{dx})^{2}]^{\frac{3}{2}}}{\frac{d^{2}y}{dx^{2}}}

               = \frac{[1 + 0.2e^{\frac{400}{1000}}^{2}]^{\frac{3}{2}}}{0.2(10^{-3})e^{\frac{400}{1000}}}

              = 3808.96 m

Now, we will calculate the normal component of the train as follows.

            a_{n} = \frac{\nu^{2}_{B}}{\rho}

                        = \frac{(26.34)^{2}}{3808.96}

                        = 0.1822 m/s^{2}

The magnitude of acceleration of train is calculated as follows.

            a = \sqrt{(a_{t})^{2} + (a_{n})^{2}}

               = \sqrt{(-0.25)^{2} + (0.1822)^{2}}

              = 0.309 m/s^{2}

Thus, we can conclude that magnitude of the acceleration of the train when it reaches point B, where sAB = 412 m is 0.309 m/s^{2}.

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Brainliest if correct Question 1 of 10
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Answer:

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