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strojnjashka [21]
3 years ago
5

There is less difference between the speed of light in glass and the speed of light in water than there is between the speed of

light in glass and the speed of light in air. Does this mean that a magnifying glass will magnify more or magnify less when it is used under water rather than in air?
Physics
1 answer:
CaHeK987 [17]3 years ago
8 0

Answer: no, the magnifying glass has nothing to do with the speed of light.

Explanation: When light moves between 2 media (refraction), part of it properties that changes during this process is it speed.

The change in speed is dependent on the refractive index of the 2 media and as given by snell's law, the refractive index is inversely proportional to wave speed.

This implies that moving from dense to a more dense medium reduces wave speed and moving from dense to less dense medium increases wave speed.

For the first statement, light moved from glass to water, it implies that it moves from a dense to a less dense medium, it wave speed increases in water.

For the second statement, light moved from glass to air, it implies that it also moves from a dense to a less dense medium and it wave speed will also increase in air.

Looking at both second medium, for the first statement, the second medium is water, and for the second statement, the second medium is air.

Air is less dense that water, hence light travel faster in air than in water.

Thus we can see that the magnification property of a glass has nothing to do with the wave speed, just the refractive indices of the media.

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5 0
3 years ago
A circular loop of flexible iron wire has an initial circumference of 165cm , but its circumference is decreasing at a constant
ArbitrLikvidat [17]

Answer:

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

Explanation:

Given data

initial circumference = 165 cm

rate = 12.0 cm/s

magnitude = 0.500 T

tome = 9 sec

to find out

emf induced and direction

solution

we know emf in loop is - d∅/dt    ........1

here ∅ = ( BAcosθ)

so we say angle is zero degree and magnetic filed is uniform here so that

emf = - d ( BAcos0) /dt

emf = - B dA /dt     ..............2

so  area will be

dA/dt = d(πr²) / dt

dA/dt = 2πr dr/dt

we know 2πr = c,

r = c/2π = 165 / 2π

r  = 26.27 cm

c is circumference so from equation 2

emf = - B 2πr dr/dt    ................3

and

here we find rate of change of radius that is

dr/dt = 12/2π = 1.91  10^{-2}cm/s

so when 9.0s have passed that radius of coil = 26.27 - 191 (9)

radius = 9.08 10^{-2} cm

so now from equation 3 we find emf

emf = - (0.500 )  2π(9.08 10^{-2} )   1.91  10^{-2}

emf = - 0.005445

and magnitude of emf = 0.005445 V

so

emf induced is 0.005445 V and direction is clockwise because we can see area is decrease and so that flux also decrease so using right hand rule direction of current here clockwise

4 0
3 years ago
GETTING TIMED PLS HELP! Do the runners in the picture above represent kinetic or potential energy? you need to explain why.
prohojiy [21]

Answer:

<em>They represent kinetic energy</em>

Explanation:

<u>Kinetic Energy </u>

A body can do work due to some of its attributes or states. For example, its mass can do work if used to provide energy, if the object is at a certain height respect to some reference level, it can do work when going downwards (potential energy), if the object moves at a certain speed, it can do work when transferring part of its speed to other objects. It's called kinetic energy and is given by

\displaystyle K=\frac{mv^2}{2}

Both runners are moving in a horizontal path, thus they have kinetic energy, given by the above equation. If they could jump below ground level, then they will also have potential energy

8 0
3 years ago
What is electron capture
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7 0
3 years ago
A jet ski accelerates towards a ramp at 2.5 m/s/s for 35 s until it finally flies off the water. Determine the
zlopas [31]

Answer:

1531 m

Explanation:

The motion of the jet ski is an uniformly accelerated motion, so we can find the distance travelled by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For the jet ski in this problem,

a=2.5 m/s^2

t = 35 s

u = 0 (it starts from rest)

Solving for s, we find the distance travelled:

d=0+\frac{1}{2}(2.5)(35)^2=1531 m

8 0
3 years ago
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