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vovikov84 [41]
4 years ago
6

Calculate the power of the eye in D when viewing an object 5.70 m away. (Assume the lens-to-retina distance is 2.00 cm. Enter yo

ur answer to at least one decimal place.)
Physics
1 answer:
Tems11 [23]4 years ago
3 0

Answer:

Power=50.17dioptre

Power=50.17D

Explanation:

P=1/f = 1/d₀ + 1/d₁

Where d₀ = the eye's lens and the object distance= 5.70m=

d₁= the eye's lens and the image distance= 0.02m

f= focal length of the lense of the eye

We know that the object can be viewed clearly by the person ,then image and lens of the eye's distance needs to be equal with the retinal and the eye lens distance and this distance is given as 0.02m

Therefore, we can calculate the power using above formula

P= 1/5.70 + 1/0.02

Power=50.17dioptre

Therefore, the power the eye's is using to see the object from distance is 5.70D

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4 years ago
A balloon has a volume of 3.5-L at 25^ * C . What would be the volume of the balloon if it were placed in a container of hot wat
Romashka-Z-Leto [24]

Answer:

4.3 L

Explanation:

Ideal gas law:

PV = nRT

Rearrange:

V / T = nR / P

Since n, R, and P are constant:

V₁ / T₁ = V₂ / T₂

Plug in values and solve:

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5 0
4 years ago
A light ray in glass (n=1.5) hits the air-glass interface at an angle of 10 degrees from the normal. What angle from the normal
jekas [21]

Answer:

The angle from the normal is 15.1°.

Explanation:

We can find the angle by using Snell's law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})

Where:

n₁: is the first medium (glass) = 1.5

n₂: is the second medium (air) = 1.0

θ₁: is the first angle (in the glass) = 10°

θ₂: is the second angle (in the air) =?

\theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.5*sin(10)}{1.0}) = 15.1 ^{\circ}

Therefore, the angle from the normal is 15.1°.

I hope it helps you!        

7 0
3 years ago
Two balls have their centers 2.0 m apart. One ball has a mass of m1 = 7.9 kg. The other has a mass of m2 = 6.1 kg. What is the g
maks197457 [2]

Answer:

3.036×10⁻¹⁰ N

Explanation:

From newton's law of universal gravitation,

F = Gm1m2/r² .............................. Equation 1

Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.

G = gravitational constant

Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²

Substituting into equation 1

F = 6.67×10⁻¹¹×7.9×6.1/2²

F = 321.427×10⁻¹¹/4

F = 30.36×10⁻¹¹

F = 3.036×10⁻¹⁰ N

Hence the force between the balls = 3.036×10⁻¹⁰ N

8 0
4 years ago
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