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worty [1.4K]
2 years ago
15

PLEASE HELP MEE I REALLY NEED IT

Physics
1 answer:
nignag [31]2 years ago
5 0

Answer:

measures of central tendency are numbers that describe what an average or typical within a distribution of data there are three main measures of central tendency mean medium and mold well they are all measuring essential tendency each is calculated differently and measures something different from others

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Which planet is the farthest?
matrenka [14]
The planet that is the farthest is Neptune, pls make me brainliest:)
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If b humans cannot see ultraviolet waves how can ultraviolet light be used to gather evidence of a crime?
Inga [223]
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You and a friend are playing with a bowling ball to demonstrate some ideas of Rotational Physics. First, though, you want to cal
RideAnS [48]

Answer:

K_{total} = 19.4 J

Explanation:

The total kinetic energy that is formed by the linear part and the rotational part is requested

         K_{total} = K_{traslation}  + K_{rotation}

let's look for each energy

linear

        K_{traslation} = ½ m v²

rotation

        K_{rotation} = ½ I w²

the moment of inertia of a solid sphere is

       I = 2/5 m r²

we substitute

       K_{total} = ½ mv² + ½ I w²

           

angular and linear velocity are related

           v = w r

we substitute

           K_{total} = ½ m w² r² + ½ (2/5 m r²) w²

           K_{total} = m w² r² (½ + 1/5)

           K_{total} = \frac{7}{10} m w² r²

let's calculate

           K_{total} = \frac{7}{10}   6.40 16.0² 0.130²

           K_{total} = 19.4 J

6 0
2 years ago
Block A of mass M is at rest and attached to the top of a spring. The block compresses the spring a distance d from its uncompre
Anni [7]

Answer:

a)  k = Mg / d , b)   v = √2gh , c)  v_{f} = \frac{2}{3} \ \sqrt{2gh},  d)   x² + 6d x - \frac{8}{3} dh = 0

e)the spring must compress a greater distance.

Explanation:

a) when the block of mass M is placed on the spring, we have an equilibrium condition,

             ∑ F  = 0

             F_{e}- W = 0

             k d = Mg

             k = Mg / d

b) let's use the concepts of energy to find the velocity of the block just before the collision

starting point. Position when released

          Em₀ = U = m g h

lowest point. Right at the point of shock

          Em_{f} = K = ½ m v²2

as there is no friction, energy is conserved

          Em₀ = Em_{f}

          mg h = ½ m v²

          v = √2gh

         

c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved

initial instant. Just before the crash

          p₀ = 2M v + M 0

final instant. Just after the shock, before the spring compression begins

         p_{f} = (2M + M) v_{f}

 the moment is preserved

          p₀ = p_{f}

          2M v = 3M v_{f}

          v_{f} = ⅔ v

          v_{f} = \frac{2}{3} \ \sqrt{2gh}

d) now we work with the joined system after the collision, let's use the concepts of energy

starting point. After shock, before beginning spring compression

        Em₀ = K = ½ (3M) v_{f}^2

        Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²

        Em₀ = 4/3 M gh

final point. With the spring fully compressed

       Em_f = K_e + U = ½ k x² + (3M) g x

in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved

         Em₀ = Em_f

        4/3 M g h = ½ k x² + 3M g x

        ½ k x² + 3Mg x - 4/3 Mgh = 0

we substitute the expression for k

         \frac{1}{2} (\frac{Mg}{d}) x² + 3Mg x - \frac{4}{3} Mgh = 0

          \frac{x^{2} }{2d} + 3 x - \frac{4}{3}h = 0

to find the value of the spring compression, the second degree equation must be solved

          x² + 6d x - \frac{8}{3} dh = 0

         x = [-6d ±\sqrt{(36 d^{2} - 4 \frac{8}{3} dh)  } ] / 2

         x = [-6d ± 6d \sqrt{ 1 -  \frac{32}{3 \ 36}  \ \frac{h}{d}    }  ]/2

         x = 3d ( -1±  \sqrt{ 1 - 0.296 \frac{h}{d}   }  )

e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.

8 0
3 years ago
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The answer to this question is D
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