Centre of mass for
(i) The equilateral triangle is the point of intersection three medians
(ii) The cylinder is the middle point of the axis of the cylinder or center of the cylinder along the axis of a cylinder.
(iii) The square is the point of intersection of two diagonals

(i) Antilons make sand traps in such a way that when an insect enters the edge of the trap, it starts to slide towards the bottom where the antilon hide itself. The angle of inclination of sand trap is made to be equal to angle of repose.
(ii) Children are fond of playing on sliding board. Sliding will be easier when the angle of inclination of the board is greater than the angle of repose. At the same time if inclination angle is much larger than the angle of repose, the slider will reach the bottom at greater speed and get hurt.

The two forces equal (i.e) P = Q
 Here R = \(\sqrt{3p}\)
R = \(\sqrt{P^2+Q^2+2PQ\cos\theta}\)
Substuting the values  \(\sqrt{3p}\) = \(\sqrt{P^2+P^2+2P^2\cos\theta}\)
3P² = P² + P² + 2P² \(\cos\theta\)
2 \(\cos\theta\) = 1
\(\cos\theta\) = \(\frac{1}{2}\)  (or) \(\theta\) = 60^o
The angle between the two forces, \(\theta\) = 60^o.

Mechanical Advantage (MA)=\(\frac { { d }_{ 2 } }{ { d }_{ 1 } } \)

\(Average\ length=\frac{2.48+1.46+2.49+2.50+2.48}{5}=\frac{12.41}{5}=2.48\ m\)
\(Mean\ absolute\ error=\frac{0.00+0.02+0.01+0.02+0.00}{5}=\frac{0.05}{5}=0.01\ m\)
\(Percentage\ error=\frac{0.01}{2.48}\times100\%=0.04\times100\%=0.40\%\)
Correct length = (2.48 ± 0.01)m
Correct length = (2.48m ± 0.40%)

\(\frac { 1 }{ 2 } m{ v }^{ 2 }\)=90% of mgh
\(\therefore\) v=6 m/s

When the block slides, the force acting on the block is kinetic friction which is equal to f_k= μ_smg.
From Newton's second law ma = -μ_smg
The negative sign implies that force acts on the opposite direction of motion.
The acceleration of the block while sliding a =-μ_kg
The negative sign implies that the acceleration is in opposite direction of the velocity. Note that the acceleration depends only on g and the coefficient of kinetic friction μ_k. We can apply the following kinematic equation
v=u+at
The final velocity is zero
0=u-u_kgt
t=\(\frac { u }{ { \mu }_{ k }g } \).

P = 8\(\times\)(kinetic energy of each bullet per second)²
= 8\(\times\)\(\frac{1}{2}\) \(\times\)(3\(\times\)10^-3)\(\times\)(600)²
P = 4320W
P = 4.320 kW

Given \(x={a^2b^3\over c\sqrt{d}}\) 
The percentage error in x is given by 
\({\triangle x\over x} \times 100=2{\triangle a \over a}\times 100+3{\triangle b\over b}\times 100 +{\triangle c\over c}\times 100+{1\over 2}{\triangle d\over d}\times 100\)
= (2\(\times\)4%)+(3\(\times\)2%)+(1\(\times\)3%)+(1\2 \(\times\) 1%)
=8% + 6% + 3% + 0.5%
The percentage error is x = 17.5%

Newton's law of gravitation states that a particle of mass M₁ attracts any other particle of mass M₂ in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.
\(\vec{F}=-\frac{G M_{1} M_{2}}{r^{2}} \hat{r}\)