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3 years ago

What affects the braking distance and thinking distance

2 answers:

8 0

Your speed is one of the only factors that has an effect on both your thinking distance and braking distance. Put simply, the faster you are going, the greater the distance travelled before you apply the brakes (thinking distance) and the vehicle comes to a complete stop (braking distance).

hodyreva [135]3 years ago

8 0

Several many things do.

Thinking distance: speed, driver's alertness or fatigue, light or dark outside, distractions inside the car, distractions outside the car, driver on the phone

Braking distance: speed, size and weight of the car or truck, condition of the tires, the road surface: wet or dry, concrete or tar, dirt or gravel

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A 500kg car is driven forward with a thrust force of 1500N. Air resistance and friction acts against the motion of the car with

Wittaler [7]

2m/s^2, this is because F=ma, meaning a is also equal to F/m. The car applies 1500N in one direction and outside sources apply a total of -500N, meaning the 500kg car is moving forward with a total of 1000N of force. Taking the total 1000N and dividing it by 500kg gives you and acceleration of 2m/s^2. Hope this helps!

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2 years ago

Which of the following best represents a chemical reaction?

Sonja [21]

Answer:

The answer to your question should be D.

Explanation:

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3 years ago

Whats the answer<br> ----------------------------

GenaCL600 [577]

Answer:

repel each other

Explanation:

The magnitude of the charge of an electron is called... ... If a positively-charged glass rod is suspended so that it turns easily and another positively-charged glass rod is brought close to it, the two rods will... Repel each other.

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3 years ago

Two small plastic spheres are given positive electrical charges. When they are a distance of 14.8cm apart, the repulsive force b

sdas [7]

Answer:

Explanation:

Case I: They have same charge.

Charge on each sphere = q

Distance between them, d = 14.8 cm = 0.148 m

Repulsive force, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times10^{9}q^{2}}{0.148^{2}}$

$q=7.56\times10^{-7}C$

Thus, the charge on each sphere is $q=7.56\times10^{-7}C$ .

Case II:

Charge on first sphere = 4q

Charge on second sphere = q

distance between them, d = 0.148 m

Force between them, F = 0.235 N

Use Coulomb's law in electrostatics

$F=\frac{Kq_{1}q_{2}}{d^{2}}$

By substituting the values

$0.235=\frac{9\times 10^{9}\times 4q^{2}}{0.148^{2}}$

$q=3.78\times10^{-7}C$

Thus, the charge on second sphere is $q=3.78\times10^{-7}C$ and the charge on first sphere is $4q = 4\times 3.78\times 10^{-7}=1.51 \times 10^{-6} C$ .

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3 years ago

Apakah yang dimaksud dengan gaya

tangare [24]

Nah gaya seperti titik fashion seperti apa yang Anda mana dan hal-hal seperti itu. (tell me if you cant understand)

8 0

3 years ago

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