Answers
Answer:
1.11m
Explanation:
From the diagram we are given the following forces;
F1 = 24.3N
F3 = 30N
Since the sum of upward forces is equal to that of downward force, then;
F2 = F1 + F3
F2 = 24.3N + 30N
F2 = 54.3N
Required
Distance between B and C
First we need to get Length of AC
Take moment about A
Anticlockwise moment = F3 cos20 * AC
Anticlockwise moment = 30ACcos 20
Clockwise moment = 1.2 * F2
Clockwise moment = 1.2(54.3) = 65.16Nm
Applying the principle of moment;
Sum of ACW moment = Sum of CW moments
30ACcos 20 = 65.16
AC = 65.16/30cos20
AC = 65.16/28.19
AC = 2.31m
Get the distance BC
AC = AB + BC
BC = AC-AB
BC = 2.31 - 1.2
BC = 1.11m
Hence the separation between B and C is 1.11m
Note that the force F1 got in (a) was the value used in the calculation.
Related Questions
Calculate the work done in lifting 200 kg of water through a vertical height of 6 metres
(g = 10 m/s)
(A) 5000 J
(B) 12000 J
(C) 25000 J
(D) 15000 J
Answers
Answer:
[tex]\bf\pink{(C)\:12000\:J}[/tex]Explanation:
Given :-[tex]\sf\red{Mass = 200 \ kg}[/tex][tex]\sf\orange{Gravity = 10 \ m/s}[/tex][tex]\sf\green{Height = 6 \ m}[/tex]Need to find :-[tex]\sf\blue{Work \ done}[/tex]Formula required :-[tex]\sf\purple{Work \ done = Mass \times Gravity \times Height}[/tex]Solution :-[tex]\to\:\:\sf\red{Work \ done = Mass \times Gravity \times Height}[/tex]
[tex]\to\:\:\sf\orange{Work \ done = 200 \times 10 \times 6}[/tex]
[tex]\to\:\:\sf\green{Work \ done = 2000 \times 6}[/tex]
[tex]\to \:\ \sf\blue{ Work \ done = {\bf{\blue{1200\:J}}}}[/tex]
Hence, [tex]\bf\green{(B)}[/tex] is the correct option.The displacement along the number line below is
a.) -6
b.) -3
c.) 0
d.) +3
e.) not shown
Answers
Answer:
d: +3
Explanation:
X1 is going +3 meters to arrive to X2. More scientifically -6+3= -3 which is the value of X2 which tells us that the correct answer is +3.