Explaining Questions - PRESSURE

1.  Explain why divers must breathe out as they rise to the surface?
   
   ANSWER
   As the diver ascends from the bottom to the surface, the pressure exerted by water decreases. This decrease in pressure causes the volume of any gas in the diver’s lungs to increase if the diver does not compensate.
2. Describe carefully how you would use a siphon to remove the major part of the water contained in a large fixed tank, open at the top. Draw a diagram and use it to explain the action of the siphon.
   ANSWER
   
   
   EXPLANATION
   Water rises in the tube because it is pushed by atmospheric pressure. The force of gravity draws water through the long arm, and this maintains the low pressure that is established at the short arm. Once start, the siphon does not require addition energy to maintain water flowing out of the tank. The siphon will pull out large part of the water in a large, fixed tank.
3.  Use a diagram to describe how manometer can be used to measure the pressure of a gas at a gas-tap. .  CONTINUE READING

Worked Examples on Archimedes Principle

Considered Formula

$$\text {i)}\quad R.D = \frac {W_a - W_l}{W_a - W_w}$$ $$\qquad \text {Where}~~R.D \to Relative~density,\quad W_a \to Weight~in~air \\\qquad W_l \to Weight~in~liquid, \qquad W_w \to Weight~in~water$$ $$\text {ii)}\quad R.D = \frac {\rho_s}{\rho_w}$$ $$\qquad \text {Where}~~R.D \to Relative~density,\quad \rho_s \to Density~of~substance \\\qquad \rho_w \to Density~of~water$$

Example 1

A piece of glass weight in air 1.2N and 0.7N when completely immersed in water calculate it’s.

a)    Relative density

b)    Density of glass

c)     Volume of glass

(Density of water = 1g/cm³)

Solution

Data given

Weight in air (W_a) = 1.2 N

Weight in water (W_w) = 0.7 N

Density of water (ρ_w) = 1g/cm³

a)        Relative density (R.D) = ?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~R.D = \frac {W_a}{W_a - W_w}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2}{1.2 - 0.7}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2}{0.5}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {1.2 \times 10}{0.5 \times 10}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac {12}{5}$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 2.4$$

          The relative density = 2.4

b)  CONTINUE READING

Effects of Forces

Forces have several effects on objects. These effects include: - 

(i)                  Stretching and Restoring force

(ii)                Torsion force

(iii)              Attraction force

(iv)              Friction force

(v)                Compression force

(vi)              Viscosity

(vii)            Repulsion force

(viii)          Air Resistance 

 

Stretching and Restoring

Stretching force is the force produce elongation of object if pulled. For Example, when spring is pulled the stretching force elongate the spring

Air Resistance 

Is the force that resists the movement of an object through the air.

Example of this .....   CONTINUE READING

Worked Examples-Calculation Part 1 - Work, Energy and Power

Considered Formula;

$$\text {i)}\quad W.d = F \times d$$ $$\qquad \text {Where}~~W.d \to Work~done,\quad F \to Force \\\qquad d \to Distance~moved$$ $$\text {ii)}\quad K.E = \frac {1}{2}mv^2$$ $$\qquad \text {Where}~~K.E \to Kinetic~energy,\quad m \to Mass \\\qquad v \to Velocity$$

Example

Abu Nufayl lifts a sack of rice by a force of 250 N against gravity to 2 m high. Calculate his work done

Solution

Data given

Force (F) =250 N

Distance (d) = 2 m

Work done (W.d) =?

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~W.d = F \times d$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 250 \times 2$$ $$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 500~Joules$$

The work done by Abu Nufayl = 500 Joules     CONTINUE READING