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4 years ago

At higher speeds, how would you compensate for the decrease in field of vision

2 answers:

5 0

When you ride a vehicle in a fast speed, then your peripheral vision will reduce that is why there is a need for you to follow the direction of the objects when you are travelling in order for you to compensate to the decrease in the field of vision.

gladu [14]4 years ago

5 0

<span>Your central and peripheral vision reduces whenever you ride a vehicle at a high speed. The best way to compensate for the decrease in your field of vision is to turn your head so that you can look to your sides. </span>

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At what point in its trajectory does a batted baseball have its minimum speed? If air resistance can be neglected, how does this

lys-0071 [83]

To understand the problem and give a logical explanation we will use the concepts related to the movement of a Projectile.

On this concept the components concerning the speed of an object that is launched in parabolic motion are characterized. There will be two types of speed components, the vertical and the horizontal.

The vertical component of velocity gradually decreases as it is affected by the force of gravity until it becomes zero at the highest point of the trajectory. After this point, it again begins to increase gradually in effect of its movement towards the gravitational attraction of the Earth.

In the case of horizontal force, if the effect of air resistance is neglected, the speed will be the same throughout the trajectory.

Now comparing this with the movement of the baseball, at the highest point, the vertical component will be zero, and the horizontal velocity will remain. Both at the beginning of the trajectory and at the end of it, there will be components of the vertical velocity, which would increase this value. Therefore the minimum speed value will be at the top, when the vertical component is zero, and the horizontal component is maintained

8 0

4 years ago

If different groups of scientist have access to the same data, how can they draw different conclusions?

Mkey [24]

I suppose they can interpret the data differently.

I hope this helps!

3 0

4 years ago

A diffraction grating, ruled with 300 lines per mm, is illuminated with a white light source at normal incidence.

Vera_Pavlovna [14]

the expression for diffraction grating allows to find the results for the questions for the angular separation are:

i) The third order is Δθ = 0.203 rad.

ii) The first order with water is Δθ = 0.046 rad.

The diffraction grating is a system formed by a large number of equally spaced lines whose diffraction is given by the expression.

d sin θ = m λ

Where d is the distance between two lines, θ is the angle of diffraction, the order of diffraction and λ is the wavelength.

i) Let's start by looking for the separation between two lines

Let's use a rule of direct proportions. If there are 300 lines in 1 mm, what distance is there between two lines.

d = 1 lines (1 mm / 300 lines) = 3,333 10⁻³ mm

d = 3.333 10⁻⁶ m

Let's find the angle of diffraction for the third order (m = 3) for each wavelength.

λ₁ = 400 nm = 400 10⁻⁹ m

sin θ₁ = $\frac{m \ \lambda }{d}$ m λ/ d

sin θ₁ = $\frac{3 \ 400 \ 10^{-9} }{3.333 \ 10^{-6} }$

θ₁ = sin⁻¹ 0.3600

θ₁ = 0.368 rad

λ₂ = 600 nm = 600 10⁻⁹ m

sin θ₂ = $\frac{3 \ 600 \ 10^{-9} }{3.333 \ 10^{-6} }$

θ₂ = sin⁻¹ 0.5401

θ₂ = 0.571 rad

The angular separation is

Δθ = θ₂ - θ₁

Δθ = 0.571 - 0.368

Δθ = 0.203 rad

ii) In this case, the separation between the network and the observation screen is filled with water.

When the rays leave the network they undergo a refraction process, for which they must comply with the relationship.

$n_i \ sin \theta_1 = n_r \ sin \theta_r$

The incident side is in the air, therefore its refractive index is n_i = 1 and when it passes into the water with refractive index n_r = 1.33.

Let's start looking for the incident angles for the first order of diffraction.

m = 1

λ₁ = 400 nm

θ₁ = sin⁻¹ $\frac{1 \ 400 \ 10^{-9}}{3.33 \ 10^{-6}}$