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:
Explanation:
Check the attachment for solution
All of the elements in a period have the same number of atomic orbitals. For example, every element in the top row (the first period) has one orbital for its electrons. All of the elements in the second row (the second period) have two orbitals for their electrons. As you move down the table, every row adds an orbital.
Answer: 0.01 m
Explanation: The formulae for capillarity rise or fall is given below as
h = (2T×cosθ)/rpg
Where θ = angle mercury made with glass = 50°
T = surface tension = 0.51 N/m
g = acceleration due gravity = 9.8 m/s²
r = radius of tube = 0.5mm = 0.0005m
p = density of mercury.
h = height of rise or fall
From the question, specific gravity of density = 13.3
Where specific gravity = density of mercury/ density of water, where density of water = 1000 kg/m³
Hence density of mercury = 13.3×1000 = 13,300 kg/m³.
By substituting parameters, we have that
h = 2×0.51×cos 50/0.0005×9.8×13,300
h = 0.6556/65.17
h = 0.01 m
