when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results
1 answer:
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Answer:
8.80 Hz
Explanation:
The frequency of a loaded spring is given by
where k and m are the spring constant and the mass of the load respectively. The values of these do not change because they are internal properties of the components of the system.
Hence, the frequency of the vertical spring mass does not change and is 8.80 Hz.
On the other hand, the frequency of the simple pendulum is affected because it is given by
where g and l are acceleration due to gravity and length of the pendulum, respectively. It is thus seen that it depends on g, which changes with location. In fact, the new frequency is given by
Answer:
1. 3
2. 1.5
3. 3 seconds
4. 0
5. 2.2 seconds
Explanation:
(1)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds
a =
(2)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=10m/s, v=22m/s and t=8 seconds
a =
(3)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
Substituting u=0m/s since at rest, v=15m/s and a=5
=
(4)
When initial and final velocity are constant, there’s no acceleration as proven below
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making a the subject we have
Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds
a =
(5)
From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.
Making t the subject we have
Substituting u=9m/s since at rest, v=0m/s and a=-4.1
=