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Vladimir79 [104]
3 years ago
12

A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/

s to 4.60 m/s in a time of 2.64 s. Calculate the tension in the string, assuming that the tension remains constant during that time.
Physics
1 answer:
Mama L [17]3 years ago
5 0

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

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Answer:

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Para levantar del suelo una caja metálica de 50 Kg se emplea maquinaria que utiliza un cable inextensible e imprime una acelerac
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Answer:

T = 530 N ,     t = 12.65 s

Explanation:

This is an exercise on Newton's second law,

           T - W = m a

where T is the tension of the cable, W the weight and the steel of the box

            T = W + ma

the weight of a body is

            W = m g

Let's replace

          T = m g + ma

          T = m (g + a)

let's calculate

          T = 50 (9.8 + 0.8)

          T = 530 N

To calculate the time we use

          y = v₀t + ½ a t²

since the box starts from rest on the sole its initial velocity is zero

          y = ½ a t2

           t = √ 2y / a

let's calculate

          t = √ (2 10 / 0.8)

         t = 12.65 s

TRASLATE

Este es un ejercicio de la segunda ley de Newton,

           T – W = m a

donde T es la tension del cable, W el peso y a la aceracion de la caja

            T = W + ma

el pesode un cuerpo es

            W = m g      

Substituyamos

          T = m g + ma

          T = m ( g +a)

calculemos

          T = 50 ( 9,8 + 0,8)

          T = 530 N

Para calcular el tiempo usamos

          y = v₀t + ½ a t²

como la caja parte del reposo en el suela su velocidad inicial es cero

          y=   ½ a t²

           t= √ 2y/a

calculemos

          t = √ (2 10/0,8)

         t = 12,65 s

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