A block attached to a spring with unknown spring constant oscillates with a period of 7.0 s. Parts a to d are independent questi
1 answer:
Answer:
a) 4.95 s
b) 7.00 s
c) 4.95 s
d) 9.9 s
Explanation:
The equation for the period of the object is
T=2×π× =
=7.0 s............................(1)
<em>m= mass of the object</em>
<em>k= spring constant</em>
<u>For part (a)</u>
m=
putting the value of m in equation (1)
T=4.95 s
<u>For part (b)</u>
the period does not depends on amplitude therefore
T=7.00 s
<u>For part (c)</u>
spring constant=2k
putting the value of spring constant in equation (1)
T=4.95 s
<u>For part (d)</u>
m= 2
putting the value of m in equation (1)
T=9.9 s
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Answer:
32.46m/s
Explanation:
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To solve this exercise we must be clear that the ball moves with constant acceleration with the value of gravity = 9.81m / S ^ 2
A body that moves with constant acceleration means that it moves in "a uniformly accelerated motion", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are the follow
Where
Vf = final speed
Vo = Initial speed =7.3m/S
A = g=acceleration =9.81m/s^2
X = displacement =51m}
solving for Vf
the speed with the ball hits the ground is 32.46m/s
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Find the acceleration using the formula a = (vf - vi) / t where:
vf = final velocity
vi = initial velocity
t = time
Plug in the points above:
vf = +10 m/s
vi = +20 m/s
t = 5 sec
(10 - 20) / 5 = -10 / 5 = -2 m/s²