False the correct answer is chemical bonds instead of thermal energy
Answer:
E = 1/2 M V^2 + 1/2 I ω^2 = 1/2 M V^2 + 1/2 I V^2 / R^2
E = 1/2 M V^2 (1 + I / (M R^2))
For a cylinder I = M R^2
For a sphere I = 2/3 M R^2
E(cylinder) = 1 + 1 = 2 omitting the 1/2 M V^2
E(sphere) = 1 + 2/3 = 1.67
E(cylinder) / E(sphere) = 2 / 1.67 = 1.2
The cylinder initially has 1.20 the energy of the sphere
The PE attained is proportional to the initial KE
H(sphere) = 2.87 / 1/2 = 2.40 m since it has less initial KE
Answer:
x=1.25m
Explanation:
The <em>Center of mass </em>of the system is defined as the point where whole mass of the body is appeared to be concentrated.
The center of mass of the system is given by
x= 
where m1 is mass of man =60 kg
m2 mass of board =20 kg
let the man be at the origin x1 =0 , x2 =5m
by substituting in above formula
x =
= 
=1.25 m
x=1.25m
So the center of mass of the system is at 1.25 m from man.
Answer:
300 J
Explanation:
First of all, we need to calculate the net force acting on the crate, which is given by:
where
F = 100 N is the horizontal push
is the force of friction
Substituting,
Now we can calculate the net work done on the crate:
where
d = 10 m is the displacement
Substituting,
According to the work-energy theorem, the kinetic energy gained by the crate is equal to the work done on it: therefore, the answer is 300 J.
Answer:
a) Maximum height reached above ground = 2.8 m
b) When he reaches maximum height he is 2 m far from end of the ramp.
Explanation:
a) We have equation of motion v²=u²+2as
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0²=5.6² + 2 x -9.81 x s
s = 1.60 m
Height above ground = 1.2 + 1.6 = 2.8 m
b) We have equation of motion v= u+at
Considering vertical motion of skateboarder.
When he reaches maximum height,
u = 6.6sin58 = 5.6 m/s
a = -9.81 m/s²
v = 0 m/s
Substituting
0= 5.6 - 9.81 x t
t = 0.57s
Now considering horizontal motion of skateboarder.
We have equation of motion s =ut + 0.5 at²
u = 6.6cos58 = 3.50 m/s
a = 0 m/s²
t = 0.57
Substituting
s =3.5 x 0.57 + 0.5 x 0 x 0.57²
s = 2 m
When he reaches maximum height he is 2 m far from end of the ramp.
