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Lady bird [3.3K]
2 years ago
10

Why is centrifugal force is a fake force?​

Physics
1 answer:
mash [69]2 years ago
5 0

Answer:

We say fictitious because the actual source of the centrifugal acceleration is somewhat indirect and the experience one has results from the unbalanced forces acting on the reference frame, not a force. Note, it is an acceleration not a force. For instance, imagine yourself on a swing.

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To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 T
Ronch [10]

Correct question is;

To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?

Answer:

We'll see more sharply in dim light

Explanation:

If we consider diffraction through a circular aperture, then angular resolution is given by;

θ = 1.22λ/D

where:

θ is the angular resolution (radians) λ is the wavelength of light

D is the diameter of the lens' aperture.

Thus,

at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm

θ = (1.22 * 550)/(2 x 10^(6))

θ = 335.5 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]

= 69.18 arc seconds

at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm

θ = (1.22 * 550)/(8 x 10^(6))

θ = 83.875 x 10^(-6) radians

Now, we need to convert this to arc seconds.

Thus;

1 arc second = 4.85 x 10^(-6) radians

So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]

= 17.3 arc seconds

From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.

4 0
3 years ago
Particles q1 = +8.0 UC, 92 = +3.5 uc, and
Olin [163]

Answer:

   F_total = 29.4 N,    directed to the right of particle 2

Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

         F = k \frac{q_1q_2}{r_{12}^2}

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

         F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

         F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

        F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

        F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

        F_total = ∑ F = F₁₂ + F₂₃

        F_total = 25.2 +3.5

        F_total = 29.4 N

directed to the right of particle 2

5 0
3 years ago
Read 2 more answers
Ocean currents are caused by water's density differences. The density differences in the ocean water are due to different salt c
kap26 [50]
The density differences in the ocean water are due to different salt concentration and differences in temperature. B) temperature
6 0
3 years ago
Read 2 more answers
A rectangular block floats in pure water with 0.400 in. above the surface and 1.60 in. below the surface. When placed in an aque
labwork [276]

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  \frac{1.60}{1.60+0.400}  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = \frac{1.60+0.400}{0.800} × 0.8

specific gravity of  solution  = 2

6 0
3 years ago
The velocity of a motor car moving along a road increases from 10m/s to 50{1} ms in 8s. Find its avarege acceleration. RESULT
lesantik [10]

Answer:

hope it helps you................

7 0
3 years ago
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