Answer:
the force of the friction is A-0.52
Answer:
400 N
Explanation:
By the law of friction,
is the maximum frictional force, 
is the coefficient of friction and 
is the reaction on the refrigerator. On a horizontal surface, the reaction is equal to the weight of the refrigerator.
While not moving, the fricition on the refrigerator is static friction. So, 
This is the maximum frictional force and is more than the applied horizontal force of 400 N. Frictional force cannot be more than the applied force, else it would actually pull the refrigerator backwards (a strange thing, if it were to happen). It is equal to the extent of the applied force because the applied force is not enough to overcome the maximum.
Hence the frictional force is 400 N.
PS: Note that we do not use the coefficient of kinetic friction because applied force could not overcome the static friction.
Answer:
A) ω = 6v/19L
B) K2/K1 = 3/19
Explanation:
Mr = Mass of rod
Mb = Mass of bullet = Mr/4
Ir = (1/3)(Mr)L²
Ib = MbRb²
Radius of rotation of bullet Rb = L/2
A) From conservation of angular momentum,
L1 = L2
(Mb)v(L/2) = (Ir+ Ib)ω2
Where Ir is moment of inertia of rod while Ib is moment of inertia of bullet.
(Mr/4)(vL/2) = [(1/3)(Mr)L² + (Mr/4)(L/2)²]ω2
(MrvL/8) = [((Mr)L²/3) + (MrL²/16)]ω2
Divide each term by Mr;
vL/8 = (L²/3 + L²/16)ω2
vL/8 = (19L²/48)ω2
Divide both sides by L to obtain;
v/8 = (19L/48)ω2
Thus;
ω2 = 48v/(19x8L) = 6v/19L
B) K1 = K1b + K1r
K1 = (1/2)(Mb)v² + Ir(w1²)
= (1/2)(Mr/4)v² + (1/3)(Mr)L²(0²)
= (1/8)(Mr)v²
K2 = (1/2)(Isys)(ω2²)
I(sys) is (Ir+ Ib). This gives us;
Isys = (19L²Mr/48)
K2 =(1/2)(19L²Mr/48)(6v/19L)²
= (1/2)(36v²Mr/(48x19)) = 3v²Mr/152
Thus, the ratio, K2/K1 =
[3v²Mr/152] / (1/8)(Mr)v² = 24/152 = 3/19
Answer: wavelength !!
hope this helped :)
To the Earth in less than ten minutes.
