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3 years ago

Four balls with different masses are dropped simultaneously from the heights

Physics

1 answer:

Scorpion4ik [409]3 years ago

7 0

D will hit last because if air resistance is null then the only force enacting on the balls is the force of gravity. Force of Gravity has an acceleration of 9.81 ms^2 and so every ball had the same acceleration and ball D is the furthest away from the floor.

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A 750 g air-track glider attached to a spring with spring constant 14.0 N/m is sitting at rest on a frictionless air track. A 20

alexandr402 [8]

Answer:

the amplitude of the subsequent oscillations is 0.11 m

the period of the subsequent oscillations is 1.94 s

Explanation:

given Information:

the mass of air-track glider, $m_{1}$ = 750 g = 0.75 kg

spring constant, k = 13.0 N/m

the mass of glider, $m_{2}$ = 200 g = 0.2 kg

the speed of glider, $v_{2}$ = 170 cm/s = 1.7 m/s

the amplitude of the subsequent oscillations is A = 0.11 m

according to mechanical enery equation, we have

$A = \sqrt{\frac{m_{1} +m_{2} }{k} }v_{f}$

where

A is the amplitude and $v_{f}$ is the final speed.

to find $v_{f}$ , we can use momentum conservation lwa, where the initial momentum is equal to the final momentum.

$P_{f} = P_{i}$

$(m_{1} +m_{2} )v_{f} = m_{1} v_{1} +m_{2}v_{2}$

$v_{1}$ = 0, thus

(0.75+0.2) $v_{f}$ = (0.75)(0)+(0.2)(1.7)

0.95 $v_{f}$ = 0.34

$v_{f}$ = 0.36 m/s

Now we can calculate the amplitude

$A = \sqrt{\frac{0.75 +0.2 }{10} }0.36$

A = 0.11 m

the period of the subsequent oscillations is T = 1.94 s

the equation for period is

T = 2π $\sqrt{\frac{m_{1}+m_{2} }{k} }$

T = 2π $\sqrt{\frac{0.75+0.2 }{10} }$

T = 1.94 s

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