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lawyer [7]
3 years ago
6

An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. the maximum speed of the o

bject is 1.25 m/s, and its maximum acceleration is 6.89 m/s2. how much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum?
Physics
1 answer:
Andreas93 [3]3 years ago
7 0
<span>First of all, the maximum speed occurs when the object passes through the
 equilibrium position

The kinetic energy when the object has this max speed is

K= 1/2 * mass * (1.25 m/s)^2

The potential energy in the spring when the speed is equal to zero

U= 1/2 * k * xmax^2

The maximun force of the spring is

mass*acceleration = k*xmax

m * 6.89 m/s2 = k * xmax
xmax = 6.89* m / k

0.5 * m * 1.56  = 0.5 * k * xmax^2

</span>m * 1.56  =  k * (<span>6.89* m / k )^2 </span>
<span>
1.56 m = 47.47 m^2 / k
m/k = 0.032862

period = 2 *pi*sqrt[m/k]
= 2 pi </span><span>sqrt [ </span><span>0.032862]
= 1.139  s

  A fourth of the period elapses between the instants of max acceleration and maximum speed

= 1/4* period
= 1/4 * </span><span><span>1.139  s </span> = 0.284s </span>






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2 years ago
A stone that is dropped freely from rest traveled half the total height in the last second. with what velocity will it strike th
alexira [117]

Answer:

hellooooo :) ur ans is 33.5 m/s

At time t, the displacement is h/2:

Δy = v₀ t + ½ at²

h/2 = 0 + ½ gt²

h = gt²

At time t+1, the displacement is h.

Δy = v₀ t + ½ at²

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h = ½ g (t + 1)²

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gt² = ½ g (t + 1)²

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(t − 1)² = 2

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Since t > 0, t = 1 + √2.  So t+1 = 2 + √2.

At that time, the speed is:

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v = g (2 + √2) + 0

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3 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
timama [110]

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

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Subtract 1 and 2 we get

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and from equation t_1-t_2=0.134\ s

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t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

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substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

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