Answer:
the planet Earth is a good example
Answer:
Explanation:
A. Using
Sinစ= y/ L = 0.013/2.7= 0.00481
စ=0.28°
B.here we use
Alpha= πsinစa/lambda
= π x (0.0351)sin(0.28)/588E-9m
= 9.1*10^-2rad
C.we use
I(စ)/Im= (sin alpha/alpha) ²
So
{= (sin0.091/0.091)²
= 3*10^-4
<span>Answer:
First we need to find the acceleration.
torque on cylinder Ď„ = T * r where T is the string tension;
T = m(g - a) where a is the acceleration of the cylinder. Then
Ď„ = m(g - a)r
But also τ = Iα. For a solid cylinder, I = ½mr²,
and if the string doesn't slip, then α = a / r, so
τ = ½mr² * a/r = ½mra.
Since Ď„ = Ď„, we have
m(g - a)r = ½mra → m, r cancel, leaving
g - a = ½a
g = 3a/2
a = 2g/3 where g, of course, is gravitational acceleration.
We know that v(t) = a*t, so for our cylinder
v(t) = 2gt / 3 â—„ linear velocity
and ω = v(t) / r = 2gt / 3r ◄ angular velocity</span>
Zeroth Law of Thermodynamics, which in turn will mean the heat will transfer between both of them up to the point where they are in thermodynamic equilibrium. So the answer is C.
