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3 years ago

A 0.500 kg mass is oscillating on a

Physics

1 answer:

drek231 [11]3 years ago

7 0

Answer:

3.6m/s

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5Kg

Spring constant (K) = 330N/m

Energy = 3.24J

Velocity (v) =..?

The velocity of the mass can be obtained as follow:

E = ½mv²

3.24 = ½ × 0.5 × v²

3.24 = 0.25 × v²

Divide both side by 0.25

v² = 3.24/0.25

Take the square root of both side

v = √(3.24/0.25)

v = 3.6m/s

Therefore, the speed of the mass is 3.6m/s

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4 years ago

Determine the internal resistance of the capacitor.

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2 years ago

What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the moveme

soldi70 [24.7K]

Answer:

The energy is $4.57x10^{-19} J$ and the wavelength is $4.34x10^{-7}m$ for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Explanation:

The energy of the absorbed photon can be known by the difference in energy between the two states in which the transition is happening (In this case from n = 2 to n = 5):

$E = E_{upper}-E_{lower}$ (1)

The permitted energy for the atom of hydrogen, according with the Bohr's model, is defined as:

$E_{n} = -\frac{13.606 eV}{n^{2}}$ (2)

Or it can be expressed in Joules, since $1eV = 1.60x10^{-19}J$

$E_{n} = -\frac{2.18x10^{-18} J}{n^{2}}$ (3)

Where the value $-2.18x10^{-18}$ represents the energy of the ground state¹ and n is the principal quantum number.

For the case of n = 2:

$E_{2} = -\frac{2.18x10^{-18} J}{(2)^{2}}$

$E_{2} = -5.45x10^{-19} J$

For the case of n = 5:

$E_{5} = -\frac{2.18x10^{-18} J}{(5)^{2}}$

$E_{5} = -8.72x10^{-20} J$

Replacing those values in equation (1) it is gotten:

$E = -8.72x10^{-20} J-(-5.45x10^{-19} J )$

$E = 4.57x10^{-19} J$

The wavelength can be determined by means of the Rydberg formula:

$\frac{1}{\lambda} = R(\frac{1}{n_{l}^{2}}-\frac{1}{n_{u}^{2}})$ (4)

Where R is the Rydberg constant, with a value of $1.097x10^{7}m^{-1}$

For this particular case $n_{l} = 2$ and $n_{u} = 5$ :

$\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(\frac{1}{(2)^{2}}-\frac{1}{(5)^{2}})$

$\frac{1}{\lambda} = 1.097x10^{7}m^{-1}(0.21)$

$\frac{1}{\lambda} = 2303700m^{-1}$

$\lambda = \frac{1}{2303700m^{-1}}$

$\lambda = 4.34x10^{-7}m$

So the energy is $4.57x10^{-19} J$ and the wavelength is $4.34x10^{-7}m$ for the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 2 to the orbit with n = 5.

<em>In what part of the electromagnetic spectrum do we find this radiation? </em>

In the Ultraviolet part of the electromagnetic spectrum.

Key terms:

¹Ground state: State of minimum energy.

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3 years ago

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OLga [1]

Answer:

Explanation:

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