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2 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because

1 answer:

7 0

Poor visibility, difficulties in colour perception, lessened colour contrast vision due to darker shadows and reduced peripheral vision, these are the reasons why one has to be extra careful while driving during hours of sunrise, sunset and night time.

<h3><u> Explanation: </u></h3>

Sunrise, sunset and night time are parts of the day with minimal or absolutely no presence of sunlight. To safely navigate roads, we require enough light in order to detect presence of other vehicles, signs and pedestrians. Less sunlight during sunrise and sunset light the sky but makes the roads and vehicles have a darker, less bright view. The contrast between colours is the least, making it difficult to identify objects and see clearly.

A rising or a setting sun can also lead to glares on the driver’s view and thus obstruct it. Since a change in ambient light is observed, our eyes need to adjust with this change and this isn’t spontaneous. Night time driving has headlight glares from approaching vehicles and reduced surrounding visibility. The eyes switching for vision adaptability from dark to bright light if vehicles approach and pass by is not a quick action. Hence the driver’s vision is compromised in every such case and this may lead to accidents.

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago

THE RELATIVE ANGLE AT THE KNEE CHANGES FROM 0O TO 85O DURING THE KNEE FLEXION PHASE OF A SQUAT EXERCISE. IF 10 COMPLETE SQUATS A

Mice21 [21]

Answer: angular distance = 1700° and 29.7 rad

also the angular displacement = 0

Explanation:

To explain this, i will give a breakdown of this works.

we are asked to find both the angular distance and displacement the knee undergo.

Ok to get the distance of the knee, we would first take note that for one to squat down and get back up, the knee would travel through 85° of flexion to gpo down, and also through another 85° of extension to return standing (upright). So, the actual angular distance of the squat is 170°.

taking ten squats, the knee would have to go through 170° motion times 10 i.e;

10 * 170° = 1700°

Therefore the angulaar distance is 1700°

now converting this distance to radians since we will be required to have our answer in both degree and rad.

Given that 2pi = 360°, it means that one degree will give 57.3°;

∅ (rad) = ∅ (deg) * 2π/360°

∅ (rad) = 1170° * 2π/360° = 29.7 rad

∅ (rad) = 29.7 rad

b. For the other part, let us remember that angular displacement is equal to angular distance divided by time, so the angular displacement displacement of the knee will be zero, because the knee's position at the final third will be the same as the initial position.

cheers i hope this helps!!!!

5 0

2 years ago

Calculate the volume of a liquid with a density of 5.45 g/ml and a mass of 65g

katrin [286]

Density=mass/volume
5.45g/ml=65g/V
V=65g/5.42g/ml
V=11.92ml

5 0

2 years ago

Raphael refers to a wave by noting its wavelength. lucinda refers to a wave by noting its frequency. which student is correct an

blsea [12.9K]

They are both right because you can note both things, I mean Raphael and Lucinda, both has a right statement or explanation about the wave. Wave by nothing is both for its wavelength and for its frequency. So Raphael and Lucinda are both correct because you can note both wavelength and frequency.

7 0

2 years ago

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

Each point is "compared to 1950-1980 baseline". So the set of data for those years is being compared to itself. No wonder it matches up pretty close !

3 0

2 years ago