Answer:
Explanation:
This problem bothers on the energy stored in a spring in relation to conservation of energy
Given data
Mass of block m =200g
To kg= 200/1000= 0.2kg
Spring constant k = 1.4kN/m
=1400N/m
Compression x= 10cm
In meter x=10/100 = 0.1m
Using energy considerations or energy conservation principles
The potential energy stored in the spring equals the kinetic energy with which the block move away from the spring
Potential Energy stored in spring
P.E=1/2kx^2
Kinetic energy of the block
K.E =1/mv^2
Where v = velocity of the block
K.E=P.E (energy consideration)
1/2kx^2=1/mv^2
Kx^2= mv^2
Solving for v we have
v^2= (kx^2)/m
v^2= (1400*0.1^2)/0.2
v^2= (14)/0.2
v^2= 70
v= √70
v= 8.36m/s
a. Distance moved if the ramp exerts no force on the block
Is
S= v^2/2gsinθ
Assuming g= 9. 81m/s^2
S= (8.36)^2/2*9.81*sin60
S= 69.88/19.62*0.866
S= 69.88/16.99
S= 4.11m
Explanation:
Below is an attachment containing the solution.
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
You can use this response or change the words around if you’d like to :)
answer: people can achieve an environmentally sustainable society by recycling, saving water, reducing the amount of household energy use and lastly reselling and donating items!
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Answer:
It is important to note, that the 2nd Law of thermodynamics plays no fundamental role in answering this question; we need a heat sink because the entropy is a state function, and at the end of the reversible process (which is visualized through the Carnot cycle diagram relevant for this problem), the entropy value of the system must return to the value it had originally.