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Gre4nikov [31]
3 years ago
10

An inattentive driver is traveling 18.0 m/s when he notices a red light ahead. his car is capable of decelerating at a rate of 3

.35 m/s2 . if it takes him 0.200 s to get the brakes on and he is 20.0 m from the intersection when he sees the light, will he be able to stop in time?
Physics
1 answer:
ryzh [129]3 years ago
3 0

Speed of the car given initially

v = 18 m/s

deceleration of the car after applying brakes will be

a = 3.35 m/s^2

Reaction time of the driver = 0.200 s

Now when he see the red light distance covered by the till he start pressing the brakes

d_1 = v* t

d_1 = 18* 0.200 = 3.6 m

Now after applying brakes the distance covered by the car before it stops is given by kinematics equation

v_f^2 - v_i^2 = 2 a s

here

vi = 18 m/s

vf = 0

a = - 3.35

so now we will have

0^2 - 18^2 = 2*(-3.35)(s)

s = 48.35 m

So total distance after which car will stop is

d = d_1 + d_2

d = 48.35 + 3.6 = 51.95 m

So car will not stop before the intersection as it is at distance 20 m

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nignag [31]

Answer:

a) fr =\mu N= 0.2*164.64 N= 32.928 N

And we need a force F> 32.928 N[tex]b) [tex] d = \frac{-v^2_i}{2a}= \frac{-(3.5m/s)^2}{-2*1.96 m/s^2}= 3.13m

Explanation:

Part a

For this case we have the following forces illustrated on the figure attached.

If we analyze on the x axis we just have two forces, fr the friction force and F the force to mantain the movement.

So we need this condition to satisfy the movement:

F > fr

If we analyze the forces on the y axis we have this:

\sum Fy= ma_y = 0

Because we have constant speed for this reason the acceleration is 0

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N = mg = 16.8Kg * 9.8 \frac{m}{s^2}= 164.64 N

And by definition the friction force is defined as: fr = \mu N

So then the friction force would be:

fr =\mu N= 0.2*164.64 N= 32.928 N

And we need a force F> 32.928 N

Part b

For this case we assume that F =0 and we have a friction force of 32.928 N.

From the second law of Newton we have:

F = ma

a = \frac{F}{m}= -\frac{fr}{m}= -\frac{32.928 N}{16.8 Kg}= -1.96m/s^2

vf =0 (final velocity, rest at the end) and vi = 3.5 m/s.

And we can find the time for the motion like this:

vf = v_i + at

t = \frac{-v_i}{a}= \frac{-3.5 m/s}{-1.96 m/s^2}= 1.78 s

And then we can find the distance from the following formula:

v^2_f= v^2_i + 2a d

d = \frac{-v^2_i}{2a}= \frac{-(3.5m/s)^2}{-2*1.96 m/s^2}= 3.13m

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The answer would be:

Move from one atom to another; the atom that looses an electron becomes positively charged and the atom that gains an electron becomes negatively charged.

It is the movement of electrons from on atom to another that enables elements to form chemical bonds. If electrons do not move, they are shared. The atom that loses an electron becomes positive because it would have less electrons than protons. The atom that accepts the electron becomes negative because it would have more electrons than protons.

Just remember that all elements are neutral, or they have the same number of protons and electrons. They become charged when they lose or gain one.

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Answer:

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Explanation:

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