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Arlecino [84]
2 years ago
6

In broad daylight, the size of your pupil is typically 3 mm. In dark situations, it expands to about 7 mm. How much more light c

an it gather?
Physics
1 answer:
Radda [10]2 years ago
3 0

Answer:

31.4 mm²

Explanation:

Light gathering power of any telescope or eye can be defined by the formula,

GDP=\pi } \frac{d^{2} }{4}

Here, d is the diameter of the pupil.

Now for bright daylight the typical size or diameter of the pupil is 3 mm.

GDP_{b} =\pi \frac{3^{2} }{4}

Now for dark situation the typical size or diameter of the pupil is 7 mm.

GDP_{b} =\pi \frac{7^{2} }{4}

Increase in light gathering power.

Increase=\pi \frac{49}{4} -\pi \frac{9}{4} \\Increase=31.4 mm^{2}

Therefore the more light which can gather by eye is 31.4 mm² .

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Assuming the starting height is 0.0 m, calculate the potential energy of the cart after it has been elevated to a height of 0.5
Bogdan [553]
The potential energy is most often referred to as the "energy at rest" and is dependent on the elevation of an object. This can be calculated through the equation,

     E = mgh

where E is the potential energy, m is the mass, g is the acceleration due to gravity, and h is the height. In this item, we are not given with the mass of the cart so we assume it to be m. The force is therefore,

   E = m(9.8 m/s²)(0.5 m) = 4.9m

Hence, the potential energy is equal to 4.9m.
8 0
2 years ago
Does a fish appear closer or farther from a person wearing swim goggles with an air pocket in front of their eyes than the fish
Sunny_sXe [5.5K]

Answer:

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

Explanation:

This exercise can be analyzed with the law of refraction that establishes that a ray of light when passing from one medium to another with a different index makes it deviate from its path,

      n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of the incident and refracted means and the angles are also for these two means.

In this case, the index of seawater replacement is 1.33, the index of refraction of air is 1, which is why the angle of replacement is less than the incident angle, so the fish seems to be closer

1 sin θ₁ = 1.33 sin θ₂

        θ₂ = sin⁻¹ ( 1/1.33 sin θ₁)

In the opposite case, when the fish looked at the face of the man, the angle of greater reason why it seems to be further away

4 0
2 years ago
For a sine wave depicting simple harmonic motion, the smaller the amplitude of the wave, the smaller the of the pendulum from th
stiks02 [169]
Displacement   , shorter 
7 0
2 years ago
Read 2 more answers
A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and
tankabanditka [31]

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

h=\dfrac{1}{2}gt_1^2

t_1=\sqrt{\dfrac{2h}{g}}

The time taken by sound =t₂

t_2=\dfrac{h}{343}

The total time

t = t₁+ t₂

6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}

6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}

Now by solving the above equation we get

h= 161.06 m

Therefore the depth of the well will be 161.06 m.

6 0
2 years ago
A father demonstrates projectile motion to his children by placing a pea on his fork's handle and rapidly depressing the curved
MariettaO [177]

Answer:

4.17 m/s

Explanation:

To solve this problem, let's start by analyzing the vertical motion of the pea.

The initial vertical velocity of the pea is

u_y = u sin \theta = (7.39)(sin 69.0^{\circ})=6.90 m/s

Now we can solve the problem by applying the suvat equation:

v_y^2-u_y^2=2as

where

v_y is the vertical velocity when the pea hits the ceiling

a=g=-9.8 m/s^2 is the acceleration of gravity

s = 1.90 is the distance from the ceiling

Solving for v_y,

v_y = \sqrt{u_y^2+2as}=\sqrt{(6.90)^2+2(-9.8)(1.90)}=3.22 m/s

Instead, the horizontal velocity remains constant during the whole motion, and it is given by

v_x = u cos \theta = (7.39)(cos 69.0^{\circ})=2.65 m/s

Therefore, the speed of the pea when it hits the ceiling is

v=\sqrt{v_x^2+v_y^2}=\sqrt{2.65^2+3.22^2}=4.17 m/s

5 0
2 years ago
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