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V125BC [204]
3 years ago
15

Lottery balls are thrown into a hopper and are set into motion by air forced through the container. each ball has a mass of 0.00

44 kg. at one instant in time, ball 1 is moving upward with a speed of 1.3 m/s, and ball 2 is moving downward with a speed of 1.66 m/s. if the kinetic energy of the three-ball system at this instant is 0.0268 j, what must be the speed of ball 3?
Physics
1 answer:
Marina CMI [18]3 years ago
3 0
In this three body system total kinetic energy remains constant:
E_{k} = constant =0.0268 J

E_{k} = E_{k1}+ E_{k2}+ E_{k3}
If we assume that upwards movement is positive and downwards movement is negative the formula is:
E_{k} = E_{k1}- E_{k2}+ E_{k3}

Formula for kinetic energy is:
E_{k} = \frac{m v^{2} }{2}

Now we have:
E_{k} = \frac{m v_{1} ^{2} }{2} -\frac{m v_{2} ^{2} }{2} +\frac{m v_{3} ^{2} }{2} \\ E_{k} = \frac{m}{2} (v_{1} ^{2}-v_{2} ^{2}+v_{3} ^{2}) \\ \\ 0.0268= \frac{0.0044}{2} (1.3 ^{2}-1.66^{2}+v_{3} ^{2}) \\ \\ 0.0536=0.0022(1.69-2.7556+v_{3} ^{2}) \\ \\ v_{3} ^{2}=25.43 \\ \\ v_{3} =+-5.04

We got two solutions: positive and negative. The speed is 5.04 m/s and direction can be upwards and downwards.
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Answer:

a)  total moment of inertia is 1359.05 kg m^2

b) angular acceleratio is 0.854rad/sec^2

Explanation:

Given data:

m1=6.9 kg

L=4.88 m

m2=34.5 kg

R=1.22 m

we klnow that moment of inertia for rod is given as

J1=(1/12) ×m×L^2

J1 = (1/12) \times 6.9 \times 4.88^2 = 13.693 kg m^2

moment of inertia for sphere is given as

J1=(2/5) ×m×r^2

J1 = (2/5) \times 34.5 \times 1.22^2 = 20.539 kg m^2

As object rotates around free end of rod then for sphere the axis around what it rotates is at a distance of d2=L+R

For rod distance is  d1=0.5*L

By Steiner theorem

for the rod we get J_1'=J_1 + m_1\times d_1^2

J_1' = 13.693 + 2.44^2\times 6.9 = 54.77 kg m^2

for the sphere we get J_2' = J_2 + m_2\times d_2^2

J2' = 20.539 + 34.5\times 6.1^2 = 1304.28 kg^m2

And the total moment of inertia for the first case is

J_{t1} = J_1'+J_2' = 54.77 + 1304.28 = 1359.05kg.m^2

b) F=476 N

The torque for system is given as

M = F\times d\times sin(a)

where a is angle between Force and distance d

and where d represent distance from rotating axis.

In this case a = 90 degree  

M = F\times L/2

M=476*2.44 = 1161.44 Nm

The acceleration is calculated as

a_1 = \frac{M}{J_{t1}}

      = \frac{1161.44}{1359.05}

      = 0.854 rad/sec^2

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Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

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3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

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final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

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        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

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          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

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