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serious [3.7K]
4 years ago
9

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

Physics
2 answers:
inysia [295]4 years ago
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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The students changed the release distance of the sphere during their investigation. Which statements best describe the purpose f
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Since unbalanced forces acts on the spheres, the aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

<h3>What is an experiment?</h3>

An experiment is a study whose objective is to study cause and effect relationships. In this case, the students had to increase the distance at which the spheres was released.

The aim of increasing the release distance of the spheres is to observe the effect of unbalanced forces acting on the spheres.

Learn more about experiments: brainly.com/question/11256472

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2 years ago
Explain what exponential growth is?
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Exponential growth is a pattern of data that shows greater increases with passing time, creating the curve of an exponential function.
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3 years ago
One end of a string 6.26 m long is moved up and down with simple harmonic motion at a frequency of 95 Hz . The waves reach the o
uranmaximum [27]

Answer:

Explanation:

Given

length of string L=6.26\ m

frequency f=95\ Hz

time taken by wave to reach at other end t=0.5\ s

speed of  wave is given by

v=\frac{length\ of\ string}{time\ taken}

v=\frac{6.26}{0.5}

v=12.52\ m/s

wavelength of is given by

\lambda =\frac{velocity}{frequency}

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\lambda =0.131\ m                          

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Read 2 more answers
The air is less dense at higher elevations, so skydivers reach a high terminal speed. The highest recorded speed for a skydiver
swat32

Answer:

the terminal velocity v_t= 202.96 m/s≅203 m/s

Explanation:

The expression for the terminal velocity

v_t= \sqrt{\frac{2mg}{\rho AC_d} }

here, C_d is the drag coefficient for the cylinder is 1.15

The surface density of the air at 20°C is

ρ_surface = 1.2041 kg/m^3

the density of air at an altitude of 39000 m

ρ= 4.3/100×39000 = 0.05177 kg/m^3

now substitute these values in equation above

we get

v_t= \sqrt{\frac{2\times90\times9.81}{0.05177\times0.72\times1.15} }

v_t= 202.96 m/s≅203 m/s

the terminal velocity v_t= 202.96 m/s≅203 m/s

5 0
4 years ago
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