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mrs_skeptik [129]
3 years ago
12

A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine

and hit a stationary green bumper car of mass 150 kg. The red, blue, and green bumper cars all combine.
A. 0.93 m/s
B 1.2 m/s
C 1.7 m/s
D 2.7 m/s
Physics
2 answers:
GuDViN [60]3 years ago
8 0

Answer:

B. 1.2m/s

Explanation:

as per the law of conservation of momentum,

initial momentum= final momentum

considering collision 1 : Mr * Vr + ( Mb * Vb) = ( Mr + Mb ) Vrb

where Vrb is the final velocity after 1st collision with which red and blue car moves.

therefore,  225* 3 + 0 = ( 225 + 180 ) Vrb

Vrb = 675/405

       = 1.67m/s

now collision 2 : ( Mr+ Mb) Vrb + MgVg= ( Mr+ Mg+ Mb ) V'

where V' is the total speed

675 + 0 = 555V'

V' = 675/555

   = 1.2 m/s

klemol [59]3 years ago
6 0
I think the answer for the question above its            b 1.2
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A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , an
kupik [55]

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

  • <em>mass of the car, m = 1000 kg</em>
  • <em>speed of the car, v = 50 km/h = 13.89 m/s</em>
  • <em>banking angle, θ = 20⁰</em>

The normal force on the car due to banking curve is calculated as follows;

Fcos(\theta) = mg

The horizontal force on the car due to the banking curve is calculated as follows;

Fsin(\theta) = \frac{mv^2}{r}

<em>Divide </em><em>the second equation by the first;</em>

\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m

Thus, the radius of the curved road at the given condition is 54.1 m.

Learn more about banking angle here: brainly.com/question/8169892

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2 years ago
According to Oxford Dictionaries, a spit take is an act of suddenly spitting out liquid one is drinking in response to something
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Answer:

The pressure is p_1 = 4051.4 \ Pa

Explanation:

From the question we are told that

     The gauge pressure at the mouth is  p_1

     The radius of the column is  r_2 =  4 \ mm  =  0.004 \ m

    The speed of the liquid outside the body is  v_2 =  3.1 \ m/s

      The area of the column is  A_2

       The area inside the mouth A_1 = 10 A_2

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=>       v_ 1 = v_2 *  \frac{A_2}{A_1}

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=>        v_ 1 = 0.31 \ m/s

So

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=>   r_1 = 10 * r_2

substituting values

        r_1 = 10 * 0.004

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Now the height of inside the mouth is  h_1 =  d =  2r_1 =  2* 0.04 =  0.08\ m

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