We have that the see-saw will be balance if a weight of 
is added to Nural's side of the see-saw
From the Question we are told that
Wan’s mass is 
Nurul’s mass is 
Generally
The Will be balance when the weight on both sides of the see-saw are equal
In conclusion
The see-saw will be balance if a weight of is added to Nural's side of the see-saw
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Pelvic girdle
In human anatomy, the pelvis is a complex of bones that connects the trunk and the legs, supports and balances the trunk, and houses and supports the intestines, the urinary bladder, and the internal organ. It is sometimes referred to as the bony pelvis or the pelvic girdle.
One is located on the left side of the body and the other is located on the right. They come together to make the pelvic girdle, a portion of the pelvis. The hip bones are attached to the upper portion of the skeleton at the sacrum.
The pelvic girdle's main function is to support the upper body's weight while seated and transfer that weight to the lower limbs while standing. For the muscles in the trunk and lower limbs, it functions as attachment point.
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Answer:
(a) t = 5.66 s
(b) t = 8 s
Explanation:
(a)
Here we will use 2nd equation of motion for angular motion:
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev)(2π rad/1 rev) = 23.25 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
23.25 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (23.25 rad)(2)/(1.45 rad/s²)
t = √(32.06 s²)
<u>t = 5.66 s</u>
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(b)For next 3.7 rev
θ = ωi t + (1/2)∝t²
where,
θ = Angular Displacement = (3.7 rev + 3.7 rev)(2π rad/1 rev) = 46.5 rad
ωi = initial angular speed = 0 rad/s
t = time = ?
∝ = angular acceleration = 1.45 rad/s²
Therefore,
46.5 rad = (0 rad/s)(t) + (1/2)(1.45 rad/s²)t²
t² = (46.5 rad)(2)/(1.45 rad/s²)
t = √(64.13 s²)
<u>t = 8 s</u>
Answer:
The diameter of the camera aperture must be greater than or equal to 1.49m
Explanation:
Let the distance separating two objects, x = 6.0 cm = 0.06 m
The distance between the observer and the two objects, d = 160 km = 160000 m
Let ∅ = minimum angular separation between the two objects that the satellite can resolve
tan( ∅) = x/d
Since there is minimum angular separation, tan( ∅) ≈∅
∅ = x/d
∅ = 0.06/160000
∅ = 3.75 * 10⁻⁷rad
For the satellite to be able to resolve the objects,
D ≥ 1.22λ/∅
λ = 560 nm = 560 * 10⁻⁹
D ≥ 1.22 * (560 * 10⁻⁹)/(3.75 * 10⁻⁷)
D ≥ 149.33 * 10⁻² m
D ≥ 1.49 m
