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3 years ago

15

Eye at the lowest radiated power of 1,2 x10 x (- 17) W. Determine how many photons of light with a wavelength of 500nm fall on t

he retina of the eye every second

1 answer:

mars1129 [50]3 years ago

8 0

Answer:

$\frac{n}{t} = 30\ photons/s$

Explanation:

The radiated power can be given in terms of the wavelength as follows:

$Rasiated\ Power = \frac{nE}{t} = \frac{nhc}{\lambda t}$

where,

Radiated Power = 1.2 x 10⁻¹⁷ W

n = no. of photons = ?

h = plank's constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

t = time

Therefore,

$1.2\ x\ 10^{-17}\ W = \frac{n(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{(5\ x\ 10^{-7}\ m)(t) }\\\\\frac{n}{t} = \frac{1.2\ x\ 10^{-17}\ W}{3.975\ x\ 10^{-19}\ J}\\\\\frac{n}{t} = 30\ photons/s$

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The muzzle velocity of an armor-piercing round fired from an M1A1 tank is 1770 m/s (nearly 4000 mph or mach 5.2). A tank is at t

loris [4]

<h2>Height of cliff is 66.43 meter.</h2>

Explanation:

Consider the horizontal motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 1770 m/s

Acceleration, a = 0 m/s²

Displacement, s = 6520 m

Substituting

s = ut + 0.5 at²

6520 = 1770 x t + 0.5 x 0 xt²

t = 3.68 s

Time of flight of shell = 3.68 s.

Now consider the vertical motion of shell,

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3.68 s

Substituting

s = ut + 0.5 at²

s = 0 x 3.68 + 0.5 x 9.81 x 3.68²

s = 66.43 m

Height of cliff is 66.43 meter.

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3 years ago

Jack and Jill are maneuvering a 3300 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is

Paraphin [41]

Answer:

The workdone by Jack is $W_{jack} = -1050J$

The workdone by Jill is $W_{Jill} = 0J$

The final velocity is $v = 1.36 m/s$

Explanation:

From the question we are given that

The mass of the boat is $m_b = 3300kg$

The initial position of the boat is $P_i = (2 \r i + 0 \r j + 3\r k)m$

The Final position of the boat is $P_f = (4\r i + 0 \r j + 2\r k )\ m$

The Force exerted by Jack $\r F = (-420\r i + 0 \r j + 210\r k) \ N$

The Force exerted by Jill $\r F_{Jill} =(180 \r i + 0\r j + 360\r k)$

Now to obtain the displacement made we are to subtract the final position from the initial position

$\r P = P_f - P_i$

$= (4\r i + 0\r j + 2 \r k) - (2\r i + 0\r j + 3\r k )$

$= (2\r i + 0\r j -\r k )m$

Now that we have obtained the displacement we can obtain the Workdone

which is mathematically represented as

$W =\r F * \r P$

The amount of workdone by jack would be

$W_{jack} =\r F * \r P$

$= [(-420\r i +0\r j +210\r k)(2\r i + 0\r j - \r k)]$

$= (-420) (2) + (210)(-1)$

$= -840 - 210$

$=-1050J$

The amount of workdone by Jill would be

$W_{Jill} =\r F * \r P$

$= [(180 \r i + 0\r j + 360\r k)(2\r i +0\r j -\r k)]$

$= (180 )(2) +(360)(-1)$

$=0J$

According to work energy theorem the Workdone is equal to the kinetic energy of the boat

$W = K.E = \frac{1}{2} m *[v^2 - (1.1)^2]$

$-1050 = 0.5*3300 [*v^2- (1.1)^2]$

$-1050 = 1650 [v^2 -1.21]$

$0.6363 = v^2 -1.21$

$v^2 = 0.6363+1.21$

$v^2 =1.846$

$v = 1.36\ m/s$

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4 years ago

The figure shows an overhead view of a ring that can rotate about its center like a merry-go-round. Its outer radius R2 is 0.8 m

Shtirlitz [24]

The cat increase the kinetic energy of the cat-ring system is 42.4 J.

Mass of the merry-go-round = M = 7.6 kg

Outer radius of the merry-go-round = R2 = 0.9 m

Inner radius of the merry-go-round = R1 = R2/2 = 0.9/2 = 0.45 m

Moment of inertia of the merry-go-round = I

I = 3.8475 kg.m2

Mass of the cat = m = M/4 = 7.6/4 = 1.9 kg

Initially the cat is sitting at the outer edge that is at a distance of 'R2' from the center.

Initial moment of inertia of the system = I1

I1 = I + mR22

I1 = 3.8475 + (1.9)(0.9)2

I1 = 5.3865 kg.m2

Initial angular speed of the system = title=View image!

1 = 7.6 rad/s

Now the cat walks to the inner edge of the merry-go-round therefore it is at a distance 'R1' from the center.

New moment of inertia of the system = I2

I2 = I + mR12

I2 = 3.8475 + (1.9)(0.45)2

I2 = 4.23225 kg.m2

New angular speed of the system =

I = I2

(5.3865)(7.6) = (4.23225)2

= 9.673 rad/s

Initial kinetic energy of the system = E1

E2 = 198 J

Amount by which the kinetic energy of the system increases =

E = E2- E1

title=View image!

E = 198 - 155.6

ΔE = 42.4 J

Amount by which the kinetic energy of the system increases when the cat crawls to the inner edge = 42.4 J.

Kinetic Energy:

Kinetic energy is the energy that an object possesses due to its motion. It is defined as the work required to accelerate an object of specified mass from rest to a specified velocity. After the body acquires this energy during acceleration, it retains this kinetic energy as long as the velocity does not change.

Learn about Kinetic Energy here:

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1 year ago

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Iteru [2.4K]

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