Answer:
The gauge pressure in Pascals inside a honey droplet is 416 Pa
Explanation:
Given;
diameter of the honey droplet, D = 0.1 cm
radius of the honey droplet, R = 0.05 cm = 0.0005 m
surface tension of honey, γ = 0.052 N/m
Apply Laplace's law for a spherical membrane with two surfaces
Gauge pressure = P₁ - P₀ = 2 (2γ / r)
Where;
P₀ is the atmospheric pressure
Gauge pressure = 4γ / r
Gauge pressure = 4 (0.052) / (0.0005)
Gauge pressure = 416 Pa
Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa
Answer:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3
Answer:
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Explanation:
The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

K= constant depends on the system of units used.
W= weight =485 g
D= density =7.9 g/cm³
A = exposed specimen area =100 in² =6.452 cm²
K=534 to give CPR in mpy
K=87.6 to give CPR in mm/yr
mpy


=37.4mpy
mm/yr


=0.952 mm/yr
Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.
Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm