Check Your UnderstandingSuppose the radius of the loop-the-loop inExample 7.9is 15 cm and thetoy car starts from rest at a heigh
1 answer:
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Answer:
change in height is 1.664 mm
Explanation:
Given data
drops = 3.00 mm
diameter = 5.00 cm = 0.05 mm
decrease = 350 cm^3
temperature = 95°C to 44.0°C
to find out
the decrease in millimeters in level
solution
we will calculate here change in volume so we can find how much level is decrease
change in volume = β v change in temp ...............1
here change in volume = area× height
so = /4 × d² h
so we can say change in volume = /4 × d² × change in height .......2
so from equation 1 and 2 we calculate change in height
( β(w) -β(g) )× v× change in temp = /4 × d² × change in height
change in height = 4 × ( β(w) -β(g) ) v× change in temp / /4 × d²
put all value here
change in height = 4 × ( 210 - 27 )(350 ) × (95-44) /
/4 × 0.05²
change in height is 1.664 mm
Answer:
33 N
Explanation:
v = Velocity of fluid = 8+2 = 10 m/s
= Density of fluid = 1.2 kg/m³
C = Coefficient of drag = 1.1
A = Cross sectional area = 0.5 m²
Drag force is given by
The drag force on the athlete is 33 N
Answer:
I will explain the concept of magnetic field and how it can be calculated.
Explanation:
The formula for magnetic field at the center of a loop is given as
B = μI / 2R
where B is the magnetic field
R is the radius of the loop
I is the current
and μ is the magnetic permeability of free space which is a constant 4π ×
newtons/ampere²
If the magnetic field at the center of the loop is 0, then μI = 0
I = 0 which means there will be no current flow in the loop.