An 800-N billboard worker stands on a 4.0-m scaffold supported by vertical ropes at each end. If the scaffold weighs 500 N and t
1 answer:
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2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north
Force in an electric field = charge * electric field
given
charge = 2* C
electric field = 20 N/C
Force in an electric field = 2* * 20
= 0.04 N
since , external field is in north direction then the force must be in north direction because the direction of an electric field at a point is the same as the direction of the electric force acting on a positive test charge
hence, 2. 0 mc charge in an external field of 20n/c, north (direction) will experience a force of 0.4 newtons, the direction of the force is north
learn more about electric field
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Answer:
f = 55mm, h ’= -9.89 cm
f = 200 mm, h ’= 42.5 cm
Explanation:
For this exercise let's start by finding the distance to the image, using the equation of the constructor
where f is the focal length, p and q are the distances to the object and image, respectively
lens with f₁ = 55mm = 0.55cm
=
= 1.718
q₁ = 0.582 m
lens with f₂ = 200mm = 2m
=
= 0.4
q₂ = 2.5 m
the magnification of a lens is given by
m =
h ’=
let's calculate for each lens
f = 55mm
h '= - 0.582 / 10 1.7
h ’= 0.0989 m
h ’= -9.89 cm
f = 200 mm
h '= - 2.5 / 10 1.7
h ’= -0.425 m
h ’= 42.5 cm
The negative sign indicates that the image is real and inverted
Answer:
The wavelength of the light decreases as it enters into the medium with the greater optical density.
The frequency of the light remains constant as it transitions between materials.
Explanation:
- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:
where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.
- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement
The frequency of the light remains constant as it transitions between materials.
is correct.
- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation
where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement
The wavelength of the light decreases as it enters into the medium with the greater optical density.
is also correct.