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Anarel [89]
3 years ago
8

A car starts from rest and speeds up at 2.2m/s^2 after the traffic light turns green , how far will it have gone when it os trav

eling at 18m/s?
PLZ write down the steps and the equation you used and thx <3
Physics
1 answer:
alex41 [277]3 years ago
3 0

73.6m

Explanation:

Initial velocity, U = 0

moving velocity, V = 18m/s

 Acceleration = 2.2m/s²

Unknown:

Distance covered = ?

Solution:

 we have to use the appropriate equation of motion to solve this problem

           V² = U² +2as

    V is the final velocity

    U is the initial velocity

     a is the acceleration

     s is the distance

  since u = 0

             V² = 2as

             18²  = 2 x 2.2 x s

            324 = 4.4s

                  s = \frac{324}{4.4} = 73.6m

Learn more:

Motion brainly.com/question/2607086

#learnwithBrainly

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shusha [124]

Answer:

μk = 0.26885

Explanation:

Conceptual analysis

We apply Newton's second law:

∑Fx = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

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m =  W/g = 75/9.8= 7.65 kg :  Block mass

Friction force : Ff

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N : Normal force (N)  

Problem development

We apply the formula (1)

∑Fy = m*ay    , ay=0

N-W-25 = 0

N = 75 +25

N= 100N

∑Fx = m*ax    

20-Ff= m*ax    

20-μk*100 = 7.65*(-0.90 )

20+7.65*(0.90) = μk*100

μk = ( 20+7.65*(0.90)) / (100)

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4 0
3 years ago
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For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

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let's find how long the ball remained in the air.

\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}

6 0
1 year ago
A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s
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To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

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On the other hand we have the conservation of the moment, which for this case would be defined as

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