53. The plan of a building is in the form of a rectangle with
centerline dimension of outer walls as 9.7mx14.7m. The
thickness of the wall in super structure is 0.30m. Then its
plinth area is
a) 150m
b) 145m2
c) 145.5m
d) 135.36m
.
Answer: 150m
Explanation:
The following can be depicted from the question:
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Therefore, the plinth area of the building will be:
= (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
= 10 × 15
= 150m
The plinth area will be 150m.
Given information
Dimensions of outer walls = 9.7m × 14.7m.
Thickness of the wall = 0.30 m
Plinth area of the building = (9.7 + 0.30/2 + 0.30/2) × (14.7 × 0.30/2 + 0.30/2)
Plinth area of the building = 10 × 15
Plinth area of the building = 150m
Therefore, the Option A is correct.
Read more about Area
brainly.com/question/25292087
When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note that the rope runs around the pulley B and a pin attached to block C, as indicated.
Answer:
The answer is "2 m/s".
Explanation:
The triangle from of the right angle:
[tex]\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant[/tex]
Differentiating the above equation:
[tex]\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\[/tex]
[tex]\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\[/tex]
[tex]= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}[/tex]