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sergejj [24]
2 years ago
7

A standard 14.16-inch (0.360-meter) computer monitor is 1024 pixels wide and 768 pixels tall. Each pixel is a square approximate

ly 281 micrometers on each side. Up close, you can see the individual pixels, but from a distance they appear to blend together and form the image on the screen.
If the maximum distance between the screen and your eyes at which you can just barely resolve two adjacent pixels is 1.30 meters, what is the effective diameter d of your pupil? Assume that the resolvability is diffraction-limited. Furthermore, use lambda = 550 nanometers as a characteristic optical wavelength.
Physics
1 answer:
Vlad1618 [11]2 years ago
4 0

Answer:

0.0031 m

Explanation:

y = Length of pixel = 281 μm

L = Distance to screen = 1.3 m

\lambda = Wavelength = 550 nm

d = Pupil diameter

\theta = Angle

We have the expression

tan\theta=\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{y}{L}\\\Rightarrow \theta=tan^{-1}\dfrac{281\times 10^{-6}}{1.3}

We have the expression

sin\theta=1.22\dfrac{\lambda}{d}\\\Rightarrow d=\dfrac{1.22\lambda}{sin\theta}\\\Rightarrow d=\dfrac{1.22\times 550\times 10^{-9}}{sin\left(tan^{-1}\frac{281\times 10^{-6}}{1.3}\right)}\\\Rightarrow d=0.0031\ m

The pupil diameter is 0.0031 m

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2 years ago
Pions have a half-life of 1.8 x 10^-8 s. A pion beam leaves an accelerator at a speed of 0.8c. What is the expected distance ove
Nuetrik [128]

Answer:

the expected distance is 4.32 m

Explanation:

given data

half life time = 1.8 × 10^{-8} s

speed = 0.8 c = 0.8 × 3 × 10^{8}

to find out

expected distance over

solution

we know c is speed of light in air is 3 × 10^{8} m/s

we calculate expected distance by given formula that is

expected distance = half life time × speed   .........1

put here all these value

expected distance = half life time × speed

expected distance = 1.8 × 10^{-8} ×  0.8 × 3 × 10^{8}

expected distance = 4.32

so the expected distance is 4.32 m

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2 years ago
If a galaxy is located 200 million light years from Earth, what can you conclude about the light from that galaxy?
natulia [17]
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8 0
2 years ago
Read 2 more answers
Suppose that, instead of the Coulomb force law, one finds experimentally that the force between any two charge q1 and q2 is Writ
denpristay [2]

Answer: E= KQ/r^2

Explanation: An electric field is a region where an electric charge(positive or negative ) will experience a force.

The magnitude of an electric field E, at a point is given by Coulombs law as

E/ F/q

Where F= Coulombs force exertedon the charge and q= electric charge

E= F/q=(KQq)/r^2q

E=KQ/r^2

6 0
2 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
2 years ago
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