Vertically, the object is in equilibrium, so that the net force in this direction is
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
where <em>n</em> is the magnitude of the normal force due to the contact between the object and surface. You're given that the object's weight is <em>mg</em> = 550 N, so <em>n</em> = 550 N as well.
Horizontally, the net force would be
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where <em>p</em> is the magnitude of the applied force and <em>f</em> is the magnitude of (kinetic) friction opposing <em>p</em>. Now,
<em>f</em> = 0.012<em>n</em> = 0.012 (550 N) = 6.6 N
so that you need to apply a force of <em>p</em> = 6.6 N to keep the object sliding at a steady pace.
Its speed decreases, its wavelength increases, and its frequency Remains the Same.
Answer:
The fraction of the ball's kinetic energy stored in the spring is 3.6KJoules
Explanation:
Given
Mass of ball =0.5kg
Velocity of ball =120m/s
The kinetic energy stored by the ball is expressed as
K.E=1/2(m*v²)
Substituting our data into the expression we have
K. E=(0.5*120²)/2
K. E=7200/2
K.E=3,600 JOULE
K.E=3.6KiloJoules
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:Reducing mass i.e. water
Explanation:
Frequency For given mass in glass is given by

where k =stiffness of the glass
m=mass of water in glass
from the above expression we can see that if mass is inversely Proportional to frequency
thus reducing mass we can increase frequency