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Oxana [17]
3 years ago
7

Which of the following sets of characteristics describes the image formed by a plane mirror?A. Virtual and invertedB. Real and u

prightC. Virtual and uprightD. Real and invertedE. All the previous statements can be correct

Physics
1 answer:
tiny-mole [99]3 years ago
4 0
<h2>Answer: Virtual and upright</h2><h2 />

A plane mirror is a highly polished flat surface with a very high capacity to reflect incident light.  

We can understand in a better way how this works with the figure attached:  

1. The incident rays coming from the real object reach the mirror and  

2.are reflected following the law of Reflection.  

3.The prolongation of those reflected rays converge at a point that does not coincide with the actual position of the object. At that point the virtual image of the object is formed.  

4.Then, the reflected divergent rays are captured by our eye converging on the retina.  

Now, the image is said to be virtual because it is a copy of the object that looks as if the object is behind the mirror and not in front of it or on the surface, but it is not really there. However, it can be seen when we focus it with our eyes.  

In addition, the image formed is:  

symmetrical, because apparently it is at the same distance from the mirror  

the same size as the object.  

upright, because it retains the same orientation as the object.

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Identify the physical property of a material that is NOT a good conductor of heat.
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3 years ago
On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the
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Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

\frac{F_E}{F_G}=\frac{qE}{mg}              (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

F_E-F_G=ma\\\\a=\frac{qE-mg}{m}

you replace the values of all variables:

a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

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3 years ago
In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that
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Answer:

On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means

\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m

This means that the relation between the wavelength and the length of the string is

3\lambda/2 = L

By definition, this standing wave is at the third harmonic, n = 3.

Furthermore, the standing wave equation is as follows:

y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)

The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.

v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)

a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0

For this equation to be equal to zero, sin(59.94t) = 0. So,

59.94t = \pi\\t = \pi/59.94 = 0.0524~s

This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:

v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s

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