Answer:
3 sec
If you are driving a 30-foot vehicle at 55 mph, how many seconds of following distance should you allow? 30ft truck. = 3 sec. Since the truck is over 40 mph.
Explanation:
Answer:
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Answer:
the third statement is true
Explanation:
given data
Lenovos cost more than Dells
Lenovos cost less than Apples
solution
we have given 1st statement that is express as
cost (Lenovo) > cost (Dell) ..................1
and
2nd statement that is express as
cost (Lenovo) < cost (Apple)
so we can say it as
cost (Apple) > cost (Lenovo) ......................2
and
now above Both equation 1 and 2 can be written as
cost (Apple) > cost (Lenovo) > cost (Dell) .........................3
so we can say cost of Apples is more than the cost of Lenovos and the cost of Dells
so as that given 3rd statement is true
Answer:
Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;
1. Agentes de arrastre de aire (AEA) o
2. Materiales poliméricos súper absorbentes
Explanation:
La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes
Ejemplos de agentes de arrastre de aire son;
Sulfonatos alcalinos
Acidos de resinas sulfonadas
Sales de ácidos grasos
Ejemplos de materiales poliméricos superabsorbentes son;
SAP0.26CT
SAP0.39PT.
Answer: downward velocity = 6.9×10^-4 cm/s
Explanation: Given that the
Diameter of the smoke = 0.05 mm = 0.05/1000 m = 5 × 10^-5 m
Where radius r = 2.5 × 10^-5 m
Density = 1200 kg/m^3
Area of a sphere = 4πr^2
A = 4 × π× (2.5 × 10^-5)^2
A = 7.8 × 10^-9 m^2
Volume V = 4/3πr^3
V = 4/3 × π × (2.5 × 10^-5)^3
V = 6.5 × 10^-14 m^3
Since density = mass/ volume
Make mass the subject of formula
Mass = density × volume
Mass = 1200 × 6.5 × 10^-14
Mass M = 7.9 × 10^-11 kg
Using the formula
V = sqrt( 2Mg/ pCA)
Where
g = 9.81 m/s^2
M = mass = 7.9 × 10^-11 kg
p = density = 1200 kg/m3
C = drag coefficient = 24
A = area = 7.8 × 10^-9m^2
V = terminal velocity
Substitute all the parameters into the formula
V = sqrt[( 2 × 7.9×10^-11 × 9.8)/(1200 × 24 × 7.8×10^-9)]
V = sqrt[ 1.54 × 10^-9/2.25×10-4]
V = 6.9×10^-6 m/s
V = 6.9 × 10^-4 cm/s
