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Kipish [7]
2 years ago
14

If the coefficient of kinetic friction between tires and dry pavement is 0.93. What is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 33.6 m/s ?
Physics
1 answer:
Alexeev081 [22]2 years ago
4 0

The shortest distance over which a car moving at 33.6 m/s may be stopped by locking the brakes is  61.93 m

The kinetic friction coefficient is 0.93.

using N, where N is the normal reaction = mg, frictional force can be evaluated.

Fr =- 0.93 x 9.8 = -9.114 m/s²

Friction, which takes the form of a decelerating force, helps the car come to a stop.

This frictional force equals the ma force, 

where ma = 7.84 m and a = 7.84 m/s2.

However, this acceleration is negative at 7.84 m/s2 since it is decelerating.

So, using the equations of motion,

v² = u² + 2ax

where;

v = final velocity

u = initial velocity

a = acceleration

0 = 33.6² + 2(-9.114)x

0 = 1128.96 - 18.228x

18.228x = 1128.96

x = 1128.96/18.228

x = 61.93 m

Therefore, the shortest distance is 61.93 m

To know more about shortest distance refer to:  brainly.com/question/16751593

#SPJ4

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Unpolarized light with intensity 370 W/m2 passes first through a polarizing filter with its axis vertical, then through a second
vampirchik [111]

To solve this problem it is necessary to apply the concepts given by Malus regarding the Intensity of light.

From the law of Malus intensity can be defined as

I' = \frac{I_0}{2} cos^2 \theta

Where

\theta =Angle From vertical of the axis of the polarizing filter

I_0 = Intensity of the unpolarized light

The expression for the intensity of the light after passing through the first filter is given by

I = \frac{I_0}{2}

Replacing we have that

I = \frac{370}{2}

I = 185W/m^2

Re-arrange the equation,

I'= \frac{I_0}{2}cos^2\theta

Re-arrange to find \theta

cos^2\theta = \frac{2I'}{I_0}

cos^2\theta = \frac{2*138}{370}

\theta = cos^{-1}(\sqrt{\frac{2*138}{370}})

\theta = 0.5282rad

\theta = 30.27\°

The value of the angle from vertical of the axis of the second polarizing filter is equal to 30.2°

4 0
3 years ago
Consider two spaceships, each traveling at 0.50c in a straight line. Ship A is moving directly away from the Sun and ship B is a
attashe74 [19]

Answer:

The velocity of the light will be 1.0c only

Explanation:

The velocity of the light measured in the case given in question will be 1.0c only.

This is due to the fact that the velocity of light is never relative. The velocity of the light is maximum

The velocity of the light cannot be scaled down in no case

Thus, the velocity of the light remains as constant.

Hence, the velocity of the light measured will be 1.0c although the ships have relative velocity.

3 0
3 years ago
What effect does temperature have on the composition of ocean water?
Burka [1]
The answer is letter B. XD
4 0
3 years ago
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A 2300 Kg car accelerates from rest to 6.00 m/s in 12.00 seconds. What is the net force acting in the car?
pav-90 [236]
F=ma
a=(v2-v1)/(t2-t1)
a=(6-0)/(12-0)
a=6/12
a= .5 m/s^2
f=2300kg*.5m/s^2
f=1150N
f=1200N if using correct sig figs
4 0
3 years ago
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Please help me solve all of them ( a, b, c and d ) thankiew !! <br> I’m also kind of in a rush
Sergio039 [100]

Answer:

a-

V= IR

9V = I ×( 12+6)

I = 9/ 18 A = 0.5 A

b

V=IR

240 = 6 A ×( 20 + R)

40 = 20 + R

R = 20 ohm

c

resultant resistance of the 2 parallel resistances= Ro

1/Ro = 1/ 5 + 1/ 20

1/Ro =( 20+5)/100

= 1/Ro = 1/4

Ro= 4 ohm

V=IR

V = 2A × ( 1+ 4 OHM)

V = 10V

d

equivalent resistance = Ro

1/Ro = 1/(2+8) + 1/(5+5)

1/Ro = 1/10 +1/10

2/10 = 1/ Ro

Ro= 10/2 = 5 ohm

V = IR

12V = I × 5Ohm

I=2.4 A

6 0
2 years ago
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