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Shalnov [3]
3 years ago
12

The speed of a vehicle is reduced with a constant acceleration from 72km/h to 18

Engineering
1 answer:
Radda [10]3 years ago
7 0

Answer:

The correct answer will be "1477.84 N".

Explanation:

Given that,

Mass,

m = 1.6 mg

or,

   = 1600 kg

Initial velocity,

u = 72 km/h

  = 72\times \frac{5}{18} \ m/s

  = 20 \ m/s

Final velocity,

v = 18 km/h

  = 18\times \frac{5}{18}

  = 5 \ m/s

Covered distance,

s = 250 m

By using the below relation, we get

⇒  v^2=u^2+2as

On putting the values, we get

⇒  (5)^2=(20)^2+2\times a\times 250

⇒      a=-0.75 \ m/s^2 (shows the deceleration)

Slope will be given as 1 in 25, then

⇒  Sin \theta=\frac{1}{25}

           \theta=2.3^{\circ}

hence,

As we know,

⇒  \Sigma F=ma

or,

⇒  Braking \ force+350-mgSin\theta=ma

⇒  Braking \ force=ma+mgSin\theta-350

On substituting all the values, we get

⇒                           =1600(0.75+1600\times 9.81 Sin(2.3^{\circ})-350

⇒                           =1477.84 \ N

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public class Main{

    public static void main(String []args){

       

       StringBuilder reversedName = new StringBuilder();

       StringBuilder subName = new StringBuilder();

       

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