Answer:
Option B and C are True
Note: The attachment below shows the force diagram
Explanation:
The weight of the two blocks acts downwards.
Let the weight of the two blocks be W. Solving for T₁ and T₂;
w = T₁/cos 60° -----(1)
w = T₂/cos 30° ----(2)
equating (1) and (2)
T₁/cos 60° = T₂/cos 30°
T₁ cos 30° = T₂ cos 60°
T₂/T₁ = cos 30°/cos 60°
T₂/T₁ =1.73
Therefore, option a is false since T₂ > T₁
Option B is true since T₁ cos 30° = T₂ cos 60°
Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.
Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.
Answer:
The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
Explanation:
Given that,
Weight Fg = mg
Acceleration = a
Tension = T
Drag force = Fa
Vertical force = L
We need to find the correct relationships
Using balance equation
In horizontally,
The acceleration is a
...(I)
In vertically,
No acceleration
Put the value of mg
....(II)
Hence, The correct relationships are T-fg=ma and L-fg=0.
(A) and (C) is correct option.
The gravitational potential energy of the brick is 25.6 J
Explanation:
The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.
Near the surface of a planet, the gravitational potential energy is given by

where
m is the mass of the object
g is the strength of the gravitational field
h is the height of the object relative to the ground
For the brick in this problem, we have:
m = 8 kg is its mass
g = 1.6 N/kg is the strenght of the gravitational field on the moon
h = 2 m is the height above the ground
Substituting, we find:

Learn more about potential energy:
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Answer:
0.087 m
Explanation:
Length of the rod, L = 1.5 m
Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.
time period, T = 3 s
the formula for the time period of the pendulum is given by
.... (1)
where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.
Moment of inertia of the rod about the centre of mass, Ic = mL²/12
By using the parallel axis theorem, the moment of inertia of the rod about the pivot is
I = Ic + md²

Substituting the values in equation (1)


12d² -26.84 d + 2.25 = 0


d = 2.15 m , 0.087 m
d cannot be more than L/2, so the value of d is 0.087 m.
Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.
Answer:
ω2 = 216.47 rad/s
Explanation:
given data
radius r1 = 460 mm
radius r2 = 46 mm
ω = 32k rad/s
solution
we know here that power generated by roller that is
power = T. ω ..............1
power = F × r × ω
and this force of roller on cylinder is equal and opposite force apply by roller
so power transfer equal in every cylinder so
( F × r1 × ω1) ÷ 2 = ( F × r2 × ω2 ) ÷ 2 ................2
so
ω2 =
ω2 = 216.47