Answer:
Explanation:
Given that
g=9.8m/s²
The spring constant is
k=50N/m
The length of the bungee cord is
Lo=32m
Height of bridge which one end of the bungee is tied is 91m
A steel ball of mass 92kg is attached to the other end of the bungee.
The potential energy(Us) of the steel ball before dropped from the bridge is given as
P.E= mgh
P.E= 92×9.8×91
P.E= 82045.6 J
Us= 82045.6 J
Potential energy)(Uc) of the cord is given as
Uc= ½ke²
Where 'e' is the extension
Then the extension is final height extended by cord minus height of cord
e=hf - hi
e=hf - 32
Uc= ½×50×(hf-32)²
Uc=25(hf-32)²
Using conservation of energy,
Then,
The potential energy of free fall equals the potential energy in string
Uc=Us
25(hf-32)²=82045.6
(hf-32)² = 82045.6/25
(hf-32)²=3281.825
Take square root of both sides
√(hf-32)²=√(3281.825)
hf-32=57.29
hf=57.29+32
hf=89.29m
We neglect the negative sign of the root because the string cannot compressed
The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time
You can't answer this question because you aren't giving the specific type of seismic waves. There is an s-wave a p-wave and an l-wave. Those are the basic waves. An S-wave cannot travel through a liquid at all. So, obviously it travels slower than any other seismic wave.
<span>It would travel faster because their speed depends on the density and composition of material that they pass through.</span>
True
False
True
My answers
Explanation:
<em>Hi</em><em>,</em><em> </em><em>there</em><em>!</em><em>!</em>
<em>Energy</em><em> </em><em>is</em><em> </em><em>defined</em><em> </em><em>as</em><em> </em><em>the</em><em> </em><em>capacity</em><em> </em><em>or</em><em> </em><em>ability</em><em> </em><em>to</em><em> </em><em>do</em><em> </em><em>work</em><em>.</em><em> </em><em>It's</em><em> </em><em>SI</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>Joule</em><em>.</em>
<em>here</em><em>,</em>
<em>Joule</em><em> </em><em>=</em><em> </em><em>(</em><em>kg</em><em>×</em><em>m</em><em>×</em><em>m</em><em>)</em><em>/</em><em>(</em><em>s</em><em>×</em><em>s</em><em>)</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>kg</em><em>×</em><em>m</em><em>^</em><em>2</em><em>/</em><em>s</em><em>^</em><em>2</em><em>.</em>
<em>Therefore</em><em>, </em><em> </em><em>the</em><em> </em><em>derived</em><em> </em><em>unit</em><em> </em><em>is</em><em> </em><em>kg</em><em>.</em><em>m</em><em>^</em><em>2</em><em> </em><em>by</em><em> </em><em>s</em><em>^</em><em>2</em><em>.</em>
<em>Hope it helps</em><em>.</em><em>.</em><em>.</em>
