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stepladder [879]
3 years ago
5

A car has a mass of 1 200 kg. A very strong weightlifter attempts, unsuccessfully, to lift the car by applying an upward force o

f 2 500 N. What is the normal force acting on the car as the weightlifter is applying this force? Be sure to include a force diagram in your answer.

Physics
1 answer:
Setler79 [48]3 years ago
4 0

Answer:check the pic

Explanation:

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Which are characteristics of the image formed by an
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The image is real.

The image is inverted.

The image is bigger than the object.

Explanation:

I really don't know why... I got this question wrong and they said this was the answer. I wish I did. Sorry.

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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.  

8 0
3 years ago
In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward
lukranit [14]

Answer:

3.8 × 10 ⁻¹⁴ m

Explanation:

The alpha particle will be deflected when its kinetic energy is equal to the potential energy

Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C

Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C

Kinetic energy of  the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) =  9.564 × 10⁻¹³

k electrostatic force constant = 9 × 10⁹ N.m²/c²

Kinetic energy = potential energy =   k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus

r = (  9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 9.564 × 10⁻¹³ = 3.8 × 10 ⁻¹⁴ m

6 0
3 years ago
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