Answer:
k = 49 N/m
Explanation:
Given that,
Mass, m = 250 g = 0.25 kg
When the mass is attached to the end of the spring, it elongates 5 cm or 0.05 m. We need to find the spring constant. Let it is k.
The force due to mass is balanced by its weight as follows :
mg=kx

So, the spring constant of the spring is 49 N/m.
The rest energy of a particle is

where

is the rest mass of the particle and c is the speed of light.
The total energy of a relativistic particle is

where v is the speed of the particle.
We want the total energy of the particle to be twice its rest energy, so that

which means:


From which we find the ratio between the speed of the particle v and the speed of light c:

So, the particle should travel at 0.87c in order to have its total energy equal to twice its rest energy.
Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation:
Periods<span> going left to right. The periodic table also has a special name </span>for<span> its vertical columns. Each column </span>is<span> called a </span>group. The elements in eachgroup<span> have the same number </span>of<span> electrons </span>in the<span> outer orbital.</span>