Answer:
1.) Magnitude = 5596 N
2.) Direction = 60 degrees
Explanation: You are given that the breakdown vehicle A is exerting a force of 4000 N at angle 45 degree to the vertical and breakdown vehicle B is exerting a force of 2000 N
Let us resolve the two forces into X and Y component
Sum of the forces in the X - component will be 4000 × cos 45 = 2828.43 N
Sum of the forces in the Y - component will be 2000 + ( 4000 × sin 45 )
= 2000 + 2828.43
= 4828.43 N
The resultant force R will be
R = sqrt ( X^2 + Y^2 )
Substitutes the forces at X component and Y component into the formula
R = sqrt ( 2828.43^2 + 4828.43^2 )
R = sqrt ( 31313752.53 )
R = 5595.87 N
The direction will be
Tan Ø = Y/X
Substitute Y and X into the formula
Tan Ø = 4828.43 / 2828.43
Tan Ø = 1.707106
Ø = tan^-1( 1.707106 )
Ø = 59.64 degree
Therefore, approximately, the magnitude and direction of the resultant force on the truck are 5596 N and 60 degree respectively.
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Answer:
74.52s
Explanation:
The solution is shown in the picture below
Work = (displacement) x (force in the direction of the displacement)
Since none of the force is in the direction of the displacement, the work it does
is zero. Something else is making the object move. It's not the 20N force.
