The work done when a spring is stretched from 0 to 40cm is 4J.
What is work done?
Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.
The work done on the spring to stretch to 40cm is,
F = kx
where F is force, k is force constant.
k = F / x = 10 N / 20 * 10^-2 m = 50 N/m
W = 0.5 * k * (x)^2
where W = work done, k = force constant.
W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.
Therefore, the work done on the spring when it is stretched to 40cm is 4J.
To learn more about work done click on the given link brainly.com/question/25573309
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solution:
y = v0t + ½at²
1150 = 79t + ½3.9t²
0 = 3.9t² + 158t - 2300
from quadratic equations and eliminating the negative answer
t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)
t = 11.37 s to engine cut-off
the velocity at that time is
v = v0 + at
v = 79 + 3.9(11.37)
v = 123.3 m/s
it rises for an additional time
v = gt
t = v/g
t = 123.3 / 9.8
t = 12.59 s
gaining more altitude
y = ½vt
y = 123.3(12.59) /2
y = 776 m
for a peak height of
y = 776 + 1150
Answer:
D. all regions of the spectrum
Explanation:
I did some research ; )
Answer:
Final velocity of NFL line backer is 16.67 m/s.
Explanation:
From the question, we have following data about the NFL line backer:
Initial Speed of line backer = Vi = 0 m/s (Since, he starts from rest)
Distance covered by NFL line backer = s = 100 m
Time taken by the NFL line backer to complete 100 m sprint = t = 12 s
Acceleration of NFL line backer during sprint = a
Final Velocity of NFL line backer = Vf = ?
First we need to find the acceleration of the NFL line backer. For that purpose we will use 2nd equation of motion:
s = (Vi)(t) + (0.5)at²
using values:
100 m = (0 m/s)(12 s) + (0.5)(a)(12 s)²
100 m/72 s² = a
a = 1.39 m/s²
Now, we use 1st equation of motion to find Vf:
Vf = Vi + at
Vf = 0 m/s + (1.39 m/s²)(12 s)
<u>Vf = 16.67 m/s</u>
