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Lemur [1.5K]
3 years ago
14

A 1200kg car moving at 25m/s has the brakes applied with a deceeration of 8.0m/s^2. How far does the car travel before it stops?

Physics
1 answer:
Lostsunrise [7]3 years ago
5 0
V=vnaut+at 0(want it to stop)=25+-8t(negative accel cuz its deceling)
-25/-8=t t=3.125 seconds or with sig figs 3.1 seconds
x=xnaut+vnaut*time+1/2*a*time^2
x(where it ends up)=0(we start distance from when it started to brake)+25*3.1+1/2*-8*3.1^2=39.06 meters or 39.1 meters 
x=0+25*3.1-.5*8*3.1^2 is the same equation as above with nothing written in. Are you in physics ap or reg? either way friend me lol(just joined website tonight)
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Answer:

1768 N

Explanation:

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F=ma

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In this problem, we have a car of mass

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3 0
3 years ago
A helium nucleus contains two protons and two neutrons. The mass of the helium nucleus is greater than the combined masses of tw
dsp73

Answer:

False

Explanation:

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If a layer was deposited but does not appear in the rock record, what has occured
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In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

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Explanation:

From the question we are told that

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The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

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