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2 years ago

When driving at night with other traffic, you should not look directly at oncoming headlights. Instead?

2 answers:

6 0

When driving at night with other traffic, you should not look directly at oncoming headlights. Instead look toward the right edge of your lane

<h3>Further explanation </h3>

To drive at night, make sure your headlights are on. Because if its on we can see it clearly and other drivers can see you. Also when driving at night, its recommended to drive slower than you would during the daytime becuse it takes longer to see potential hazards on the road. If you feel yourself getting drowsy, pull over at a rest stop then try to get some sleep.

If an approaching car is using its high-beams, don't look directly into the oncoming headlights, instead look towards the right edge of your lane. Watch the oncoming car out of the corner of your eye. Do not try to retaliate against the other driver by keeping your high-beam lights on. If you do, both of you can be blinded. There are no permanent eye damage from looking at car headlights, but by looking directly at any light will cause discomfort and hinder vision temporarily.

<h3>Learn more</h3>

Learn more about driving at night brainly.com/question/4637543
Learn more about traffic brainly.com/question/11606596
Learn more about oncoming headlights brainly.com/question/12961845

<h3>Answer details</h3>

Grade: 9

Subject: physics

Chapter: traffic

Keywords: oncoming headlights, traffic, driving at night, look, driving

12345 [234]2 years ago

5 0

<span>
watch the right edge of your lane and check the position of oncoming cars. </span>

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3 years ago

3. 3.0*1015 electrons are removed from one side of a parallel plate capacitor and travel to the other side of the capacitor. You

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Answer:

True

$Q = 480 \mu C$

Explanation:

As we know that total charge is always quantized and in the terms of multiples of charge of single electron

so we have

Q = Ne

here we know that

$N = 3.0 \times 10^{15}$

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3 years ago

Consider a light, single-engine airplane such as the Piper Super Cub. If the maximum gross weight of the airplane is 7780 N, the

viva [34]

Answer:

V=19.08 m/s

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Air density p=1.2250 kg/m³

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Solution

The equation to find stalling speed is given below

$W=(1/2)S_{area}(V_{Stalling-speed} )^{2} (P_{Air-density} )(C_{Lmax} )\\ so\\V_{Stalling-speed}=\sqrt{\frac{2W}{S_{area}*(P_{Air-density} )(C_{Lmax} )} }\\V_{Stalling-speed}=\sqrt{\frac{2*7780}{16.6*2.1*1.225} }\\ V_{Stalling-speed}=19.08 m/s$

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3 years ago

A spherical drop of water carrying a charge of 43 pC has a potential of 540 V at its surface (with V = 0 at infinity). (a) What

almond37 [142]

Explanation:

Given that,

Charge on a spherical drop of water is 43 pC

The potential at its surface is 540 V

(a) The electric potential on the surface is given by :

$V=\dfrac{kq}{r}$

r is the radius of the drop

$r=\dfrac{kq}{V}\\\\r=\dfrac{9\times 10^9\times 43\times 10^{-12}}{540}\\\\r=7.17\times 10^{-4}\ m$

(b) Let R is the radius of the spherical drop, when two such drops of the same charge and radius combine to form a single spherical drop. ATQ,

$\dfrac{4}{3}\pi r^3+\dfrac{4}{3}\pi r^3=\dfrac{4}{3}\pi R^3\\\\2r^3=R^3\\\\R=2^{1/3} r$

Now the charge on the new drop is 2q. New potential is given by :

$V=\dfrac{9\times 10^9\times 43\times 10^{-12}\times 2}{2^{1/3}\times 7.17\times 10^{-4}}\\\\V=856.79\ V$

Hence, the radius of the drop is $7.17\times 10^{-4}\ m$ and the potential at the surface of the new drop is 856.79 V.

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2 years ago