Answer:
A star uses fusion as an energy source by building larger atoms from smaller atoms.
Explanation:
Nuclear fission and fusion are two processes at which an atomic nucleus is changed to produce energy. Fission is the process splitting heavy atomic into lighter atomic nuclei.
So, fusion is the combination of smaller atoms to form larger atoms and star uses this as source of energy.
Fusion is the process at which light atomic nuclei are merged or fused together to form heavier nuclei.
The energy source for all stars is nuclear fusion. In a nuclear fusion reaction, the nuclei of two atoms combine to create a new atom. Most commonly, in the core of a star, two hydrogen atoms fuse to become a helium atom.
Answer:
49 N
Explanation:
In order to move the box at constant speed, the acceleration of the box must be zero (a=0): this means, according to Newton's second law,
F = ma
that the net force acting on the box, F, must be zero as well.
Here there are two forces acting on the box in the horizontal direction while it is moving:
- The force of push applied by the guy, F
- The frictional force, 
For an object moving on a flat surface, the frictional force is given by

where
is the coefficient of friction
m is the mass of the box
g is the acceleration of gravity
So the equation of the forces becomes

And substituting:

We find the force that must be applied by the guy:

I’d say two soccer balls that are touching each other, I hope that helps!
(a) 
The moment of inertia of a uniform-density disk is given by

where
M is the mass of the disk
R is its radius
In this problem,
M = 16 kg is the mass of the disk
R = 0.19 m is the radius
Substituting into the equation, we find

(b) 142.5 J
The rotational kinetic energy of the disk is given by

where
I is the moment of inertia
is the angular velocity
We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

And so, the rotational kinetic energy is

(c) 
The rotational angular momentum of the disk is given by

where
I is the moment of inertia
is the angular velocity
Substituting the values found in the previous parts of the problem, we find

Answer:
Magnitude of the force is

direction of the force is given as
West of South
Explanation:
As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy
So here we have

here we know that first force is of magnitude 2 N towards east

second force is also of 2.0 N due North

now from above equation


so magnitude of the force is given as


direction of the force is given as


West of South