Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
8861.75 m approximately 8862 m
Explanation:
We need to remember Newton's 2nd Law which says that the force experienced by an object is proportional to his acceleration and that the constant of proportionality between those two vectors correspond to the mass of the object.
for the weight of an object (which is a force) we have that the acceleration experienced by that object is equal to the gravitational acceleration, obtaining that 
For simplicity we work with 
despiting the effect of the height above sea level. In this problem, we've been asked by the height above sea level that makes the weight of an object 0.30% more lighter.
In accord with the formula 
the "normal" or "standard" weight of an object is given by 
when 
, so we need to find the value of 
that makes 
meaning that the original weight decrease by a 0.30%, so now we operate...
now we group like terms on the same sides 
we cancel equal tems on both sides and obtain that 
Answer:
hello your question is incomplete attached below is the missing equation related to the question
answer : 40.389° , 38.987° , 38° , 39.869° , 40.265°
Explanation:
<u>Determine the friction angle at each depth</u>
attached below is the detailed solution
To calculate the vertical stress = depth * unit weight of sand
also inverse of Tan = Tan^-1
also qc is in Mpa while σ0 is in kPa
Friction angle at each depth
2 meters = 40.389°
3.5 meters = 38.987°
5 meters = 38.022°
6.5 meters = 39.869°
8 meters = 40.265°
Answer:
The differences are listed below
Explanation:
The differences between consolidation and compaction are as follows:
In compaction the mechanical pressure is used to compress the soil. In consolidation, there is an application of stead pressure.
In compaction, there is a dynamic load by rapid mechanical methods like tamping, rolling, etc. In consolidation, there is static and sustained pressure applied for a long time.
In compaction, the soil volume is reduced by removing air from the void. In consolidation, the soil volume is reduced by squeezing out water from the pores.
Compaction is used for sandy soil, consolidation on the other hand, is used for clay soil.
Answer:
50°
Explanation:
Complementary angles add up to 90°.
Supplementary angles add up to 180°.
Vertical angles are equal.
A + B = 90°
B = C
C = 180° − 140°
C = 40°
B = 40°
A = 50°
