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vovangra [49]
3 years ago
12

Consider two identical springs. At the start of an experiment, Spring A is already stretched out 3 cm, while Spring B remains at

the zero position. Both springs are then stretched an additional three centimeters. What conclusion can you draw about the force required to stretch these springs during the experiment
Physics
1 answer:
aksik [14]3 years ago
7 0

Answer:

Explanation:

In this interesting exercise we have that spring A is 3 cm longer, due to previous experiments if these experiments did not reach the non-linear elongation point, the cosecant Km of the spring must remain the same, therefore when we lengthen the two springs these the longitudinal are lengthened.

As a consequence of the above according to Hockey law, the prediction of lengthening is the same, therefore the outside is the same in two two systems

            F = K Δx

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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
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Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

x = 26.16 m

Part d)

y = 7.49 m

Part e)

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Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

x = v_x t

x = (15.3)(1.03)

x = 15.76 m

Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

y = 7.94 m

Part c)

Horizontal displacement in 1.71 s

x = v_x t

x = (15.3)(1.71)

x = 26.16 m

Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

y = 7.49 m

Part e)

Horizontal displacement in 5.44 s

x = v_x t

x = (15.3)(5.44)

x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(5.44) - 4.9(5.44)^2

y = -75.6 m

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