Answer:
a-1 Graph is attached. The relation is linear.
a-2 The corresponding height for 68 kPa Pressure is 7.54 m
a-3 The corresponding weight for 68 kPa Pressure is 1394726kg
b The original height of the column is 5.98 m
Explanation:
Part a
a-1
The graph is attached with the solution. The relation is linear as indicated by the line.
a-2
By the equation

Here
- P is the pressure which is given as 68 kPa.
- ρ is the density of the oil whose SG is 0.92. It is calculated as

- g is the gravitational constant whose value is 9.8 m/s^2
- h is the height which is to be calculated

So the height of column is 7.54m
a-3
By the relation of volume and density

Here
- ρ is the density of the oil which is 920 kg/m^3
- V is the volume of cylinder with diameter 16m calculated as follows

Mass is given as

So the mass of oil leading to 68kPa is 1394726kg
Part b
Pressure variation is given as

Now corrected pressure is as

Finding the value of height for this corrected pressure as

The original height of column is 5.98m
Answer:
a block sliding down a ramp,a leaf blowing across a field
Answer:
Radio Wave
Micro Wave
Explanation:
Electromagnetic waves are transverse waves composed by the perpendicular oscillating electric and magnetic fields.
EM waves have both Electrical and magnetic features.
they travel in the velocity of light (3*10⁸ ms⁻¹)
Electromagnetic spectrum is obtained according to their wave length and the frequency. Due to wave length range it's categorized. Here is the decreasing order of wave length and increasing order of different wave types in electromagnetic spectrum
- Radio Wave
- Micro Wave
- IR wave
- Light Wave
- UV rays
- X rays
- Gamma rays
- Cosmic rays
1) In a circular motion, the angular displacement

is given by

where S is the arc length and r is the radius. The problem says that the truck drove for 2600 m, so this corresponds to the total arc length covered by the tire:

. Using the information about the radius,

, we find the total angular displacement:

2) If we put larger tires, with radius

, the angular displacement will be smaller. We can see this by using the same formula. In fact, this time we have:
Using K.E=1/2MV^2
answer is 125joules