The answer would be letter choice B
Answer:
Yosef hypothesis could be stretching of rubber band depends on rubber band's width. It is difficult to stretch a wider rubber band in comparison to a narrow band.
Explanation:
Width of rubber band affect how easily it can be stretched. It is found that it is difficult to to stretch a wider rubber band in comparison to a narrow band because when rubber band is narrow less molecules will be there along its width and hence less restoring force will be there so rubber can be easily stretched. On the other hand when the rubber band is wider it means more molecules are there along its width and hence more restoring force will be there while stretching so it will be difficult to stretch a wider rubber band.
Answer:
0.018kg
Explanation:
Q=mlf
m=?,Lf=334000J/kg,Q=5848J
5848=mx334000
5848=334000m
divide both sides by 334000
334000m/334000=5848/334000
m=0.018kg
Answer:
n = 1810
A = 25 mm
Explanation:
Given:
Lateral force amplitude, F = 25 N
Frequency, f = 1 Hz
mass of the bridge, m = 2000 kg/m
Span, L = 144 m
Amplitude of the oscillation, A = 75 mm = 0.075 m
time, t = 6T
now,
Amplitude as a function of time is given as:
or amplitude for unforce oscillation
or
or
Now, provided in the question Amplitude of the driven oscillation
the value of the maximum amplitude is obtained 
thus,
Now, for n people on the bridge
Fmax = nF
thus,
max amplitude
or
n = 1810
hence, there were 1810 people on the bridge
b)
since the effect of damping in the millenium bridge is 3 times
thus,
b=3b
therefore,
or
or
or
A = 0.025 m = 25 mm
Answer:
Final velocity of the first person is 3.43m/s and that of the second person is 0.0242m/s
Explanation:
Let the momentum of the first person, the ball second person be Ma, Mb and Mc.
From the principle of the conservation of momentum, sum of the momentum before collision is equal to the sum of the momentum after collision.
Ma1 + Mb1 = Ma2 + Mb2.
The ball and the first person are both moving together with a common velocity 3.45m/s.
Let the velocity of the first person be v1
Therefore
67.5×3.45+ 0.041×3.45= 67.5v1 + 0.041×34
233.02 = 1.39+ 67.5v1
67.5v1 = 233.02 - 1.39 = 231.61
v1 = 231.61 / 67.5
v1 = 3.43m/s
The second person and the ball move together with a common velocity after catching the ball.
For the second person and the ball let their final common velocity be v
Mb2 + Mc2 = Mb3 + Mc3
0.041 × 34 + 57.5 ×0 = (57.5 + 0.041)×v
57.541v = 1.39
v = 1.39 /57.541
v = 0.0242m/s
