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
==>
Answer: KE = 62.5J
Explanation:
Given that
Mass of object = 5kg
kinetic energy KE = ?
velocity of object = 5m/s
Since kinetic energy is the energy possessed by a moving object, and it depends on the mass (m) of the object and the velocity (v) by which it moves. Therefore, the object has kinetic energy.
i.e K.E = 1/2mv^2
KE = 1/2 x 5kg x (5m/s)^2
KE = 0.5 x 5 x 25
KE = 62.5J
Thus, the object has 62.5 joules of kinetic energy.
U=10 m/s
v=30 m/s
t=6 sec
therefore, a=(v-u)/t
=(30-10)/6
=(10/3) ms^-2
now, displacement=ut+0.5*a*t^2
=60+ 0.5*(10/3)*36
=120 m
And you can solve it in another way:
v^2=u^2+2as
or, s=(v^2-u^2)/2a
=(900-100)/6.6666666.......
=120 m
Respuesta:En la astenosfera existen lentos movimientos de convección que explican la deriva continental. Además, el basalto de la astenosfera fluye por extrusión a lo largo de las dorsales oceánicas, lo cual hace que se renueve y expanda constantemente el fondo oceánico. :D
Answer:
The frequencies are 
Explanation:
From the question we are told that
The length of the ear canal is 
The speed of sound is assumed to be 
Now taking look at a typical ear canal we see that we assume it is a closed pipe
Now the fundamental harmonics for the pipe(ear canal) is mathematically represented as
substituting values
Also the the second harmonic for the pipe (ear canal) is mathematically represented as
substituting values
Given that sound would be loudest in the pipe at the frequency, it implies that the child will have an increased audible sensitivity at this frequencies
