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jasenka [17]
1 year ago
8

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° w

ith respect to the vertical.
The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground. At what distance x will the ball land?

Physics
1 answer:
notka56 [123]1 year ago
6 0

The distance x will the ball land after flies off with a horizontal initial velocity  is 3.0635 m.

<h3>What is mechanical energy?</h3>

The mechanical energy is the sum of kinetic energy and the potential energy of an object at any instant of time.

M.E = KE +PE

A small metal ball with a mass of m = 62.0 g is attached to a string of length l = 1.85 m. It is held at an angle of θ = 48.5° with respect to the vertical.

The ball is then released. When the rope is vertical, the ball collides head-on and perfectly elastically with an identical ball originally at rest. This second ball flies off with a horizontal initial velocity from a height of h = 3.76 m, and then later it hits the ground.

The conservation of energy principle states that total mechanical energy remains conserved in all situations where there is no external force acting on the system.

Kinetic energy  = Potential energy

1/2 mv² =mgh₁

The velocity at the bottom, when the height h = 5m, is

v= √2gh₁...................(1)

The vertical height h₁ = l- lcosθ

h₁ = l- lcosθ

h₁ = 1.85 - 1.85cos48.5°

h₁ =0.6241 m

Putting the values in equation (1), we get

v = √2x 9.81 x0.6241

v = 3.499 m/s

The horizontal distance traveled is

x = vt

x = v x √2h/g

Plug the values, we get

x =  3.499 x √2x3.76 / 9.81

x = 3.0635 m

Thus, the horizontal distance ball travels is  3.0635 m.

Learn more about mechanical energy.

brainly.com/question/13552918

#SPJ1

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Answer:

The distance is  s=  30.3 \ m

Explanation:

From the question we are told that

   The  coefficient of static friction is  \mu_s  =  0.42

    The  initial speed of the train is  u =  57 \  km /hr = 15.8 \ m/s

   

For the crate not to slide the friction force must be equal to the force acting on the train i.e

       -F_f  =  F

The negative sign shows that the two forces are acting in opposite direction

=>   mg  *  \mu_s  =  ma

=>   -g  *  \mu_s  = a

=>   a =  -9.8 *  0.420

=>   a =  -4.116 m/s^2

From equation of motion

  v^2  = u^2  +  2as

Here  v =  0 m/s since it came to a stop

=>   s=  \frac{v^2 - u^2 }{ 2 a}

 =>   s=  \frac{0 -(15.8)^2 }{ - 2 * 4.116}

=>     s=  30.3 \ m

7 0
3 years ago
The Assignment: A fixed quantity of an ideal gas (R 0.28 kJ/kgK; Cv-0.71kJ/kgK) is expanded from an initial condition of 35 bar,
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Explanation:

The index of expansion is the relationship of pressures:

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p1*v1/T1 = p2*v2/T2

p2 = p1*v1*T2/(T2*v2)

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The index of expansion then is 35/7.1 = 4.93

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7.1 bar = 710000 Pa

q = p2*v2 - p1*v1 + Δu

q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg

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Answer:

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\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

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  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

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