Answer:
The submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³
Explanation:
Given;
wet density of soil sample = 2.5 Mg/m³ = 25 kN/m³
Specific gravity of solid particle = 2.7
The dry unit weight of soil;
for undisturbed state, the volume of the soil is;
Submerged effective density is given as;
density of water (ρw) = 2.7 x 25 kN/m³ = 67.5 kN/m³, substitute this in the above equation;
Therefore, the submerged effective density is 86.93 kN/m³ or 8.693 Mg/m³
Answer:
Dear user,
Answer to your query is provided below
Acceleration is zero because of no change in velocity.
Explanation:
Remember that velocity is a vector quantity and a vector can change in 3 ways
•Magnitude only
•Direction only
•Both magnitude and direction.
Now the magnitude of velocity (speed) can stay constant while the direction is changing. This is the case in circular motion.
In the question above, it is mentioned that the girl is moving along a straight road. Therefore no change in direction of velocity.
All of Dina's potential energy Ep is converted into kinetic energy Ek so Ep=Ek, where Ep=m*g*h and Ek=(1/2)*m*v². m is the mass of Dina, h is the height of ski slope, g=9.8 m/s² and v is the maximal velocity.
So we solve for v:
m*g*h=(1/2)*m*v², masses cancel out,
g*h=(1/2)*v², we multiply by 2,
2*g*h=v² and take the square root to get v
√(2*g*h)=v, we plug in the numbers and get:
v=9.9 m/s.
So Dina's maximum velocity on the bottom of the ski slope is v=9.9 m/s.
Answer:
Explanation:
Let electric potential at A ,B and C be Va , Vb and Vc respectively.
Work done = charge x potential difference
Wab = q ( Va - Vb )
Wac = q ( Va - Vc )
Given
Wac = - Wab / 3
3Wac = - Wab
Now
Wbc = q ( Vb - Vc )
= q [ ( Va-Vc ) - ( Va - Vb )]
= Wac - Wab
= Wac + 3Wac
= 4Wac
