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Lostsunrise [7]
3 years ago
6

Vision. When light having vibrations with angular frequency ranging from 2.7×1015rad/s to 4.7×1015rad/s strikes the retina of th

e eye, it stimulates the receptor cells there and is perceived as visible light.
What are the limits of the frequency of this light?
Physics
1 answer:
Amiraneli [1.4K]3 years ago
6 0

Answer:

Explanation:

Given

angular frequency ranging From 2.7\times 10^{15}\ rad/s\ to\ 4.7\times 10^{15}\ rad/s

we know \omega =2\pi f

where f=frequency

\omega =angular\ Frequency

for \omega =2.7\times 10^{15}

f=\frac{2.7\times 10^{15}}{2\pi }

f=4.29\times 10^{14} Hz

for \omega =4.7\times 10^{15}

f=\frac{4.7\times 10^{15}}{2\pi }

f=7.47\times 10^{14} Hz

Frequency ranges from  4.29\times 10^{14}\ Hz\ to\ 7.47\times 10^{14} Hz          

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It is claimed that if a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all its kinet
TEA [102]

Answer:

v=346.05\ m.s^{-1}

Explanation:

Given:

initial temperature of the lead bullet, T_i=43^{\circ}C

latent heat of fusion of lead, L_f=2.32\times 10^4\ J.kg^{-1}

melting point of lead, T_m=327.3^{\circ}C

We have:

specific heat capacity of lead, c=129\ J.kg^{-1}.K^{-1}

<em>According to question the whole kinetic energy gets converted into heat which establishes the relation:</em>

\rm KE=(heat\ of\ rising\ the\ temperature\ from\ 43\ to\ 327.3\ degree\ C)+(heat\ of\ melting)

\frac{1}{2} m.v^2=m.c.\Delta T+m.L_f

\frac{1}{2} m.v^2=m(c.\Delta T+L_f)

\frac{v^2}{2} =129\times(327.3-43)+23200

v=346.05\ m.s^{-1}

3 0
3 years ago
The different between distance and displacement
balandron [24]

distance is metres in any direction ... displacement is distance in a given direction ... 1 mile versus 1 mile North

4 0
3 years ago
A motorcycle accelerated from 10 metre per second to 30 metre per second in 6 seconds. How far did it travel in this time ?
Ber [7]
U=10 m/s
v=30 m/s
t=6 sec

therefore, a=(v-u)/t
                   =(30-10)/6
                   =(10/3) ms^-2

now, displacement=ut+0.5*a*t^2
                              =60+ 0.5*(10/3)*36
                              =120 m
And you can solve it in another way:

v^2=u^2+2as
or, s=(v^2-u^2)/2a
       =(900-100)/6.6666666.......
       =120 m

7 0
4 years ago
I need help asap !
Elodia [21]

Answer:

the answer is 100%!

your welcome

5 0
3 years ago
A physics demo launches a ball horizontally while dropping a second ball vertically at exactly the same time. 1) which ball hits
Simora [160]

Answer:

Both balls will hit the ground at the same time

Explanation:

The factor which leads to ball falling is the gravity acting on the ball;

The motions along the path of both balls are independent and both balls will obey the following illustration

Using the third equation of motion

s = ut + ½at²

Where s = distance covered by both balls.

u = initial velocity of both balls. Since both balls start from rest, u = 0m/s

a = acceleration; and it's equal to acceleration due to gravity.

a = g

By substituton

s = 0 * t + ½gt²

s = 0 + ½gt²

s = ½gt²

Make t the subject of formula

gt² = 2s

t² = 2s/g

t = ±√(2s/g)

But time can't be less than 0 (in other words, negative)

So,

t = √(2s/g)

It'll take both balls √(2s/g) time to hit the floor

6 0
3 years ago
Read 2 more answers
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