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garri49 [273]
3 years ago
13

A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d

irection at 1.40 m/s. All cars have a mass of mass 1.6 104 kg. Determine the following.(a) speed of the three coupled cars after the collision (Give your answer to at least 2 decimal places.) m/s(b) kinetic energy lost in the collision J
Physics
1 answer:
I am Lyosha [343]3 years ago
6 0

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

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A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When
11111nata11111 [884]

Answer:

The value is R_f =  \frac{4}{5}  R

Explanation:

From the question we are told that

   The  initial velocity of the  proton is v_o

    At a distance R from the nucleus the velocity is  v_1 =  \frac{1}{2}  v_o

    The  velocity considered is  v_2 =  \frac{1}{4}  v_o

Generally considering from initial position to a position of  distance R  from the nucleus

 Generally from the law of energy conservation we have that  

       \Delta  K  =  \Delta P

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

      \Delta K  =  K__{R}} -  K_i

=>    \Delta K  =  \frac{1}{2}  *  m  *  v_1^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * (\frac{1}{2} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K  =  \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R  from the nucleus , this is mathematically represented as

          \Delta P =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P =  k  *  \frac{q_1 * q_2 }{R}  - 0

So

           \frac{1}{2}  *  m  * \frac{1}{4} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R}  - 0

=>        \frac{1}{2}  *  m  *v_0^2 [ \frac{1}{4} -1 ]  =   k  *  \frac{q_1 * q_2 }{R}

=>        - \frac{3}{8}  *  m  *v_0^2  =   k  *  \frac{q_1 * q_2 }{R} ---(1 )

Generally considering from initial position to a position of  distance R_f  from the nucleus

Here R_f represented the distance of the proton from the nucleus where the velocity is  \frac{1}{4} v_o

     Generally from the law of energy conservation we have that  

       \Delta  K_f  =  \Delta P_f

Here \Delta K is the change in kinetic energy from initial position to a  position of  distance R  from the nucleus  , this is mathematically represented as

      \Delta K_f   =  K_f -  K_i

=>    \Delta K_f  =  \frac{1}{2}  *  m  *  v_2^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * (\frac{1}{4} * v_o )^2  -  \frac{1}{2}  *  m  *  v_o^2

=>    \Delta K_f  =  \frac{1}{2}  *  m  * \frac{1}{16} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2

And  \Delta  P is the change in electric potential energy  from initial position to a  position of  distance R_f  from the nucleus , this is mathematically represented as

          \Delta P_f  =  P_f - P_i

Here  P_i is zero because the electric potential energy at the initial stage is  zero  so

             \Delta P_f  =  k  *  \frac{q_1 * q_2 }{R_f }  - 0      

So

          \frac{1}{2}  *  m  * \frac{1}{8} * v_o ^2  -  \frac{1}{2}  *  m  *  v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f }

=>        \frac{1}{2}  *  m  *v_o^2 [-\frac{15}{16} ]  =   k  *  \frac{q_1 * q_2 }{R_f }

=>        - \frac{15}{32}  *  m  *v_o^2 =   k  *  \frac{q_1 * q_2 }{R_f } ---(2)

Divide equation 2  by equation 1

              \frac{- \frac{15}{32}  *  m  *v_o^2 }{- \frac{3}{8}  *  m  *v_0^2  } }   =  \frac{k  *  \frac{q_1 * q_2 }{R_f } }{k  *  \frac{q_1 * q_2 }{R } }}

=>           -\frac{15}{32 } *  -\frac{8}{3}   =  \frac{R}{R_f}

=>           \frac{5}{4}  =  \frac{R}{R_f}

=>             R_f =  \frac{4}{5}  R

   

7 0
3 years ago
at the sewing store, ava bought a bag of buttons. 21 in all. 21 of the buttons were large. what percentage of the buttons wre la
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If she only has 21 buttons and all 21 of them are large, then all of her buttons are large. so 100% of the buttons would be large.
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Answer:

v = 24 m/s, rightwards

Explanation:

Given that,

The mass of TBT explosive = 5 kg

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The law of conservation of momentum holds here. Let v be the final speed of the remaining piece. So,

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babunello [35]

Answer:

True

Explanation:

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Therefore, pressure is inversely proportional to the area of the surface:

p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

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First law of motion: A rocket being launched up in the atmosphere.

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