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Alex
3 years ago
7

In a Broadway performance, an 84.5-kg actor swings from a R = 4.30-m-long cable that is horizontal when he starts. At the bottom

of his arc, he picks up his 55.0-kg costar in an inelastic collision. What maximum height do they reach after their upward swing?
Physics
1 answer:
Arada [10]3 years ago
3 0

Answer:

1.57772 m

Explanation:

M = Mass of actor = 84.5 kg

m = Mass of costar = 55 kg

v = Velocity of costar

V  = Velocity of actor

h_i = Intial height of actor = 4.3 m

g = Acceleration due to gravity = 9.81 m/s²

As the energy of the system is conserved

\frac{1}{2}MV^2=Mgh_i\\\Rightarrow V=\sqrt{2gh_i}\\\Rightarrow V=\sqrt{2\times 9.81\times 4.3}\\\Rightarrow V=9.18509\ m/s

As the linear momentum is conserved

MV=(m+M)v\\\Rightarrow v=\frac{MV}{m+M}\\\Rightarrow V=\frac{84.5\times 9.18509}{84.5+55}\\\Rightarrow v=5.56372\ m/s

Applying conservation of energy again

\frac{1}{2}(m+M)v^2=(m+M)gh_f\\\Rightarrow h_f=\frac{v^2}{2g}\\\Rightarrow h_f=\frac{5.56372^2}{2\times 9.81}\\\Rightarrow h_f=1.57772\ m

The maximum height they reach is 1.57772 m

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kompoz [17]
     The working of a spring is given by:

T_{el}= \frac{k\Delta x^2}{2}
 
     Entering the unknowns, we have:

T_{el}=\frac{k\Delta x^2}{2} \\ 1.25*10^{-2}*2*J=k(2.5*10^{-2}m)^2 \\ k= \frac{2.5*10^{-2}*J}{(2.5*10^{-2}m)^2} \\ k=40* \frac{J}{m^2} \\ k=40 \frac{N*m}{m^2}  \\ \boxed {k=40* \frac{N}{m} }
 
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4 0
3 years ago
28. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32° ramp at a sp
Ad libitum [116K]

Answer:

a) # buses = 7  

Explanation:

For this exercise we use the kinematic equations, let's find the time it takes to reach the same height

   

     y =v_{oy}  t - ½ g t²

Let's decompose the speed, with trigonometry

      v₀ₓ = v₀ cos θ

      v_{oy} = v₀ sin  θ

      v₀ₓ = 40 cos 32

      v₀ₓ = 33.9 m / s

      v_{oy} = 40 sin32

      v_{oy} = 21.2 m / s

When it arrives it is at the same initial height y = 0

         0 = (v_{oy} - ½ gt) t

That has two solutions

       t = 0                    when it comes out

       t = 2 v_{oy} / g       when it arrives

       t = 2 21.2 /9.8

       t = 4,326 s

We use the horizontal displacement equation

       x = vox t

       x = 33.9   4.326

       x = 146.7 m

To find the number of buses we can use a direct proportions rule

    # buses = 146.7 / 20

    # buses = 7.3

    # buses = 7

The distance of the seven buses is

     L = 20 * 7 = 140 m

b) let's look for the scope for this jump

     R = vo2 sin2T / g

     R = 40 2 without 2 32 /9.8

     R = 146.7 m

As we can see the range and distance needed to pass the seven (7) buses is different there is a margin of error of 6.7 m in favor of the jumper (security)

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Answer:

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Explanation:

Since the original count rate is 600 Bq,

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ii) after 2 half lives, the count rate decreases to 1/4 of 600 = 150 Bq

iii) after 3 half lives, the count rate decreases to 1/6 of 600 = 100 Bq

iv) after 4 half lives, the count rate decreases to 1/8 of 600 = 75 Bq

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As the temperature decreases, the atoms lose energy. The atoms begin to move slower. They are held together by attractive forces
bogdanovich [222]

gas to liquid is the answer

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Answer:

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