Answer:
An emergency kit
Explanation:
The reason I say this is because:
A first aid kit can aid you if you have..
Scars
Cuts
Bruises
So, i would say that The first aid kit is the life line.
Answer:
Q = 63,827.5 W
Explanation:
Given:-
- The dimensions of plate A = ( 10 mm x 1 m )
- The fluid comes at T_sat , 1 atm.
- The surface temperature, T_s = 75°C
Find:-
Determine the total condensation rate of water vapor onto the front surface of a vertical plate
Solution:-
- Assuming drop-wise condensation the heat transfer coefficient for water is given by Griffith's empirical relation for T_sat = 100°C.
h = 255,310 W /m^2.K
- The rate of condensation (Q) is given by Newton's cooling law:
Q = h*As*( T_sat - Ts )
Q = (255,310)*( 0.01*1)*( 100 - 75 )
Q = 63,827.5 W
Answer:
6.99 x 10⁻³ m³ / s
Explanation:
Th e pressure difference at the two ends of the delivery pipe due to atmospheric pressure and water column will cause flow of water.
h = difference in the height of water column at two ends of delivery pipe
6 - 1 = 5 m
Velocity of flow of water
v = √2gh
= √ (2 x 9.8 x 5)
= 9.9 m /s
Volume of water flowing per unit time
velocity x cross sectional area
= 9.9 x 3.14 x .015²
= 6.99 x 10⁻³ m³ / s
Answer:
s= 20.4 m
Explanation:
First lets write down equations for each ball:
s=so+vo*t+1/2a_c*t^2
for ball A:
s_a=30+5*t+1/2*9.81*t^2
for ball B:
s_b=20*t-1/2*9.81*t^2
to find time deeded to pass we just put that
s_a = s_b
30+5*t-4.91*t^2=20*t-4.9*t^2
t=2 s
now we just have to put that time in any of those equations an get distance from the ground:
s = 30 + 5*2 -1/2*9.81 *2^2
s= 20.4 m
Answer:
a) A suspended floor is a ground floor with a void underneath the structure. The floor can be formed in various ways, using timber joists, precast concrete panels, block and beam system or cast in-situ with reinforced concrete. However, the floor structure is supported by external and internal walls.
b) Soil exploration consists of determining the profile of the natural soil deposits at the site, taking the soil samples and determining the engineering properties of soils using laboratory tests as well as in-situ testing methods
c) Bulking in sand Occurs When dry sand interacts with the atmospheric moisture. Presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles to move aside to each other. This results in the increase of the volume of sand.
d) In a nutshell, bearing capacity is the capacity of soil to support the loads that are applied to the ground above. It depends primarily on the type of soil, its shear strength and its density. It also depends on the depth of embedment of the load – the deeper it is founded, the greater the bearing capacity.
Explanation:
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