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
We have that the speed of a body covering a distance of 320 km in 4h is mathematically given as
V=22.22m/s is
<h3 /><h3>From the question we are told
calculate the speed of a body covering a distance of 320 km in 4h
Generally the equation for the Speed is mathematically given as
V=22.22m/s
Hence
The speed of a body covering a distance of 320 km in 4h is
V=22.22m/s
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Answer:
The magnitud of the force is 124.8N.
Explanation:
First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:
μs
and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:
replacing this we have:
μs
μs*
and μs
now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.
the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.
now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.
again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:
This is the new external F force that we are looking for:
we know Friction2(AB) because we found that in the previous block so:
*μs
for the other friction we have to use the equation:
*μs
from y axis we have:
we found the value of Normal(AB) with the previous block so:
and:
*μs
*μs
*μs+μs*
*μs
and since: μs*
the new F force would be:
*μs
