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slavikrds [6]
3 years ago
12

Tests reveal that a normal driver takes about 0.75s before he or she can react to a situation to avoid a collision. It takes abo

ut 3s for a driver having 0.1% alcohol in his system to do the same.
If such drivers are traveling on a straight road at 30mph (44 )ft/s and their cars can decelerate at 24ft/s^2 , determine the shortest stopping distance d for normal driver from the moment he or she see the pedestrians.

Also, Determine the shortest stopping distance for drunk driver from the moment he or she see the pedestrians.
Physics
1 answer:
netineya [11]3 years ago
6 0

Answer:

<u>NORMAL DRIVER: </u>d = 73.3 ft

<u>DRUNK DRIVER: </u>d = 172.3

Explanation:

<u>NORMAL DRIVER:</u>

Distance covered in initial 0.75s = 0.75s *44 =  33ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 33 + 40.3 = 73.3 ft

<u>DRUNK DRIVER:</u>

Distance covered in initial 3s = 3s *44 =  132 ft

USING THE THIRD EQUATION OF MOTION

V^2-U^2 = 2as

0-(44)^2 = 2 (-24) s

s = 1936/48 =40.3 ft

d = 132 + 40.3 = 172.3 ft

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Hope you could get an idea from here.

Doubt clarification - use comment section.

8 0
3 years ago
What type of circuit is illustrated PLEASE HELPPP
kumpel [21]

Well the shown circuit is definitely open because the lines are not connected. And it is not a parallel circuit because there would be more lines. Like a string of Christmas lights. So it must be an "open series circuit".

7 0
3 years ago
Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
scZoUnD [109]

Answer:

20 cm

Explanation:

Te electric potential enery U = kq₁q₂/r  were q₁ = 5 nC = 5 × 10⁻⁹ C and q₂ = -2 nC = -2 × 10⁻⁹ C and r = √(x - 2)² + (0 - 0)² +(0 - 0)² = x - 2. U =  -0.5 µJ = -0.5 × 10⁻⁶ J, k = 9 × 10⁹ Nm²/C².

So r = kq₁q₂/U

x - 2 = kq₁q₂/U

x = 0.02 + kq₁q₂/U m

x = 0.02 + 9 × 10⁹ Nm²/C² × 5 × 10⁻⁹ C × -2 × 10⁻⁹ C/-0.5 × 10⁻⁶ J

x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

x = 0.02 + 0.18 = 0.2 m = 20 cm

7 0
3 years ago
Review. A bead slides without friction around a loop-the loop (Fig. P8.5). The bead is released from rest at a height h=3.50R (b
oksian1 [2.3K]

The normal force on the bead at point A is 1 N.

According to the Law Of Conservation of Energy, the total energy of the bead at the topmost point of height of 3.50 R is the same as its total energy at point A.

The energy at a height of 3.50R is equal to its potential energy which is

U= mgh= mg x 3.5R= 3.5mgR .........(i)

The energy at point A is,

E= mg(2R)+ 1/2 mv^2  ...........(ii)

Now, equating (i) and (ii)

3.5mgR = 2mgR + 1/2 mv^2

3.5mgR - 2mgR = 1/2mv^2

1.5gR = 1/2 v^2

3gR = v^2

V = √3gR

At point A, three forces are there weight of the bead(mg), normal reaction(N), and centripetal force(mv^2/r).

Balancing these forces,

mg + N = mv^2/r

N= mv^2/r - mg

Putting the given values in the above equation, we get the normal reaction

N= 1 N

Thus, the normal force on the bead at point A is 1N.

To know more about "Conservation of Mechanical Energy", refer to the following link:

brainly.com/question/11264649?referrer=searchResults

#SPJ4

8 0
1 year ago
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