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4 years ago

10

What happens to incoming light rays that are parallel to the principal axis of a convex lens?

2 answers:

katrin [286]4 years ago

4 0

Answer is A. They converge at the focal point

11Alexandr11 [23.1K]4 years ago

3 0

Explanation:

The imaginary line passing through the center of the lens is called the principal axis of the lens.

If the incoming light rays comes parallel to the principal axis of a convex lens, then after refraction it will pass through the principal focus on the other side of the lens. All the rays meet at the focal point.

The rays of light converge after refraction converge at a point and that point is called as the principal force of the lens.

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A DC current of 60 mA can cause paralysis of the body's respiratory muscles and hence interfere with breathing, but only 15 mA (

Liula [17]

To solve this problem it is necessary to apply the concepts related to Ohm's Law to find the voltage in the case of direct current. While for the Alternate current we use the concept of RMS voltage.

PART A) In the case of DC we have to,

V=IR

Where,

I = Current

R = Resistance

Where the current is 60mA y the resistance is $1000\Omega$ , then the potential would be

$V=(60mA)(1100\Omega)$

$V = 66V$

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$I_{rms} = 15 mA$

$R = 1100 \Omega$

$V_{rms} = 16.5 V$

The potential in AC then is,

$V_o = \sqrt{2}V_{rms}$

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$V_o = 23.3345 V$

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How far from the nucleus in angstroms (1 Å = 10-10 m) is the electron in a hydrogen atom if it has an energy of -8.72 * 10-20 J?

olganol [36]

Answer:

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Question 10 of 10

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Protons in an accelerator at the Fermi National Laboratory near chicago are accelerated to a total energy that is 400 times thei

Thepotemich [5.8K]

(a) 0.99999687c

The relationship between total energy (E) and rest energy ( $E_0$ ) is

$E=\gamma E_0$

where $\gamma$ is the relativistic factor. In this problem, we know that the total energy is 400 times the rest energy: this means that

$\gamma=400$

The formula for the relativistic factor is

$\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

where v is the speed of the proton and c is the speed of light. By isolating the term v/c, we find the speed of the protons in terms of c:

$1-(\frac{v}{c})^2=\frac{1}{\gamma^2}\\\frac{v}{c}=\sqrt{1-\frac{1}{\gamma^2}}=\sqrt{1-\frac{1}{400^2}}=0.99999687$

(b) $3.75\cdot 10^5 MeV$

Given the rest mass of a proton:

$m_0=1.67\cdot 10^{-27}kg$

its rest energy is the energy equivalent to this mass:

$E_0=m_0 c^2 = (1.67\cdot 10^{-27}kg)(3\cdot 10^8 m/s)^2=1.5\cdot 10^{-10} J$

The protons in the Fermi Laboratory have energy that is 400 times their rest energy, so their total energy is

$E=400 E_0$

The total energy is also sum of rest energy and kinetic energy:

$E=E_0+K$

So the kinetic energy is

$K=E-E_0=400E_0 - E_0 = 399 E_0 = 399(1.5\cdot 10^{-10}J)=6.0\cdot 10^{-8} J$

And since the conversion factor is

$1 eV = 1.6\cdot 10^{-19} J$

This kinetic energy converted into MeV ( $10^6 eV$ ) is

$K=\frac{6.0\cdot 10^{-8}J}{1.6\cdot 10^{-19}J/eV \cdot 10^6 eV/MeV}=3.75\cdot 10^5 MeV$

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3 years ago