To solve this problem we will apply the concept of magnification, which is given as the relationship between the focal length of the eyepieces and the focal length of the objective. This relationship can be expressed mathematically as,
Here,
= Magnification
= Focal length eyepieces
= Focal length of the Objective
Rearranging to find the focal length of the objective
Replacing with our values
Therefore the focal length of th eobjective lenses is 27.75cm
Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite
Answer:
Pressure,P=6×10^3Pa
Explanation:
The gas has an ideal gas behaviour and ideal gas equation
PV=NKT
T= V/N p/K ...eq1
Average transitional kinetic energy Ktr=1.8×10-23J
Ktr=3/2KT
T=2/3Ktr/K....eq2
Equating eq1 and 2
V/N p/K = 2/3Ktr/K
Cancelling K on both sides
P= 2/3N/V( Ktr)
Substituting the value of N/V and dividing by 10^-6 to convert cm^3 to m^3
P = 2/3 (5.0×10^20)/10^-6 × 1.8×10^-23
P= 6 ×10^3Pa
Speed = (distance traveled) / (time to travel the distance).
Strange as it may seem, 'velocity' is completely different.
Velocity doesn't involve the total distance traveled at all.
Instead, 'velocity' is based on 'displacement' ... the distance
between the start-point and end-point, regardless of the route
taken to get there. So the displacement in driving once around
any closed path is zero, because you end up where you started.
Velocity =
(displacement during some time)
divided by
(time for the displacement)
AND the direction from the start-point to the end-point.
For the guy who drove 15 km to his destination in 10 min, and then
back to his starting point in 5 min, (assuming he returned by way of
the same 15-km route):
Speed = (15km + 15km) / (10min + 5min) = (30/15) (km/min)
= 2 km/min.
Velocity = (end location - start position) / (15 min) = Zero .
