For a human jumper to reach a height of 110 cm, the person will need to leave the ground at a speed of 4.65 m/s.
We can calculate the initial speed to reach 110 cm of height with the following equation:

Where:
: is the final speed = 0 (at the maximum height of 110 cm)
: is the initial speed =?
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 110 cm = 1.10 m
Hence, the <u>initial velocity</u> is:

Therefore, the initial speed that the person must have to reach 110 cm is 4.65 m/s.
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The release of free energy drives the spontaneous reaction.
Spontaneity can be <span>determined
using the change in </span>Gibbs free energy
(the thermodynamic potencial):
delta G=delta H – T*delta
S
where delta H is the enthalpy and delta S is the entropy.
The direction (the sign) of delta G depends of the changes
of enthalpy and entropy. If delta G is negative then the process is
spontaneous.
In our case, both delta H and delta S are negative values, the
process as said is spontaneous which means that it may proceed in the forward
direction.
The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>