Answer:
the correct distance is 202 ft
Explanation:
The computation of the correct distance is shown below:
But before that correction to be applied should be determined
= (101 ft - 100 ft) ÷ (100 ft) × 200 ft
= 2 ft
Now the correct distance is
= 200 ft + 2 ft
= 202 ft
Hence, the correct distance is 202 ft
The same would be relevant and considered too
Answer:
a) the object floats
b) the object floats
c) the object sinks
Explanation:
when an object is less dense than in the fluid in which it is immersed, it will float due to its weight and volume characteristics, so to solve this problem we must find the mass and volume of each object in order to calculate the density and compare it with that of water
a)
volumen for a cube
V=L^3
L=1.53in=0.0388m
V=0.0388 ^3=5.8691x10^-5m^3=58.69ml
density=m/v
density=13.5g/58.69ml=0.23 g/ml
The wooden block floats because it is less dense than water
b)
m=111mg=0.111g
density=m/v
density=0.111g/0.296ml=0.375g/ml
the metal paperclip floats because it is less dense than water
c)
V=0.93cups=220.0271ml
m=0.88lb=399.1613g
Density=m/v
density=399.1613/220.027ml=1.8141g/ml
the apple sinks because it is denser than water
Answer:A
Explanation:
Damp roof is generally applied at basement level which restrict the movement of moisture through walls and floors. Therefore it could be inside or the outside basement walls.
Answer:
point B where 
has the largest Q value at section a–a
Explanation:
The missing diagram that is suppose to be attached to this question can be found in the attached file below.
So from the given information ;we are to determine the point that has the largest Q value at section a–a
In order to do that; we will work hand in hand with the image attached below.
From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.
We also have block partitioned into different point segments . i,e A,B,C, D
For point A ;
Let Q be the moment of the Area A;
SO ; 
where ;
For point B ;
Let Q be the moment of the Area B;
SO ; 
where ;
For point C ;
Let Q be the moment of the Area C;
SO ; 
where ;
For point D ;
Let Q be the moment of the Area D;
SO ; 
since there is no area about point D
Area = 0
Thus; from the foregoing ; point B where has the largest Q value at section a–a
