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Svetradugi [14.3K]
3 years ago
9

Desde una altura de 120 m se deja caer un cuerpo. Calcular a los 2,5 s i) la rapidez que lleva; ii) cuánto ha descendido; iii) c

uánto le falta por
Physics
1 answer:
shtirl [24]3 years ago
5 0

Responder: A.) 24.5m / s B.) 30.625m C.) 89.375m

Explicación:

Dado lo siguiente:

Altura desde la cual se cae el cuerpo = 120 m

Tiempo (t) = 2.5s

A.) La velocidad que toma:

El cuerpo cayó desde una altura;

velocidad inicial (u) = 0

Para calcular v:

V = u + en

Donde a = aceleración debido a la gravedad = 9.8m / s

v = 0 + (9.8) (2.5)

v = 24.5 m / s

B) Cuánto ha disminuido.

Usando la ecuación de movimiento:

S = ut + 0.5at ^ 2

Donde S = distancia

S = 0 × 2.5 + 0.5 (9.8) (2.5 ^ 2)

S = 0 + 0.5 (9.8) (6.25)

S = 30.625 m

Esta es la distancia recorrida después de 2.5 segundos Altura o distancia ha disminuido en 30.625 m

C.) ¿CUÁNTO FALTA? Por lo tanto, 120m - 30.625m = 89.375m

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2 years ago
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

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Explanation:

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        Em₀ = U = mgh

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This is the velocity of the body when it reaches the ground, so the force remains

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where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

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2 years ago
Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k
natima [27]

Answer:

part (a) a_1\ =\ 2.9\ kg

Part (b) a_2\ =\ 6.25\ kg

Explanation:

Given,

  • mass of the smaller disk = M_1\ =\ 0.900\ kg
  • Radius of the smaller disk = R_1\ =\ 2.45\ cm\ =\ 0.0245\ m
  • mass of the larger disk = M_2\ =\ 1.6\ kg
  • Radius of the larger disk =R_2\ =\ 5.0\ cm\ =\ 0.05\ m
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Let I be the moment of inertia of the both disk after the welding,\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2

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From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)

Net torque on the smaller disk,

\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)

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