Answer:
Answered
Explanation:
A) The work done by gravity is zero because displacement and the gravitational force are perpendicular to each other.
W= FS cosθ
θ= 90 ⇒cos90 = 0 ⇒W= 0
B) work done by tension
W= Tcosθ×S= 5cos30×2.30= 10J
C) Work done by friction force
W= f×s=1×2.30= 2.30 J
D) Work done by normal force is Zero because the displacement and the normal force are perpendicular to each other.
E) The net work done= Work done by tension in the rope - frictional work
=10-2.30= 7.7 J
The stage where atoms are spread out and bouncy is the gas stage.
Answer
given,
mass of copper rod = 1 kg
horizontal rails = 1 m
Current (I) = 50 A
coefficient of static friction = 0.6
magnetic force acting on a current carrying wire is
F = B i L
Rod is not necessarily vertical
the normal reaction N = mg-F y
static friction f = μ_s (mg-F y )
horizontal acceleration is zero
B_w = B sinθ
B_d = B cosθ
iLB cosθ= μ_s (mg- iLB sinθ)
B = 0.1 T
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
Answer:
The final angular speed is 16.1 rad/s
Explanation:
Given;
initial moment of inertia, I₁ = 2.56 kg.m²
final moment of inertia, I₂ = 0.40 kg.m²
initial angular speed, ω₁ = 0.4 rev/s = 2.514 rad/s
Apply the principle of conservation of angular momentum;
I₁ω₁ = I₂ω₂
where;
ω₂ is the final angular speed
ω₂ = (I₁ω₁) / (I₂)
ω₂ = (2.56 x 2.514) / (0.4)
ω₂ = 16.1 rad/s
Therefore, the final angular speed is 16.1 rad/s
