Answer:
0.6375 m/s
Explanation:
Let x be the distance of the man from the building
from the figure attached
initially the value of x=12
Given:

where the negative sign depicts that the distance of the man from the building is decreasing.
Now, Let The length of the shadow be = y
we have to calculate
when x=4
from the similar triangles
we have,
or

Differentiating with respect to time 't' we get

or

Now for x = 4, and
we have,

or

<u>here, the negative sign depicts the decrease in length and in the question it is asked the decreasing rate thus, the answer is </u><u>0.6375m/s</u>
Answer:
The acceleration of man 1 and 2 is
and
.
Explanation:
Mass of man 1, m₁ = 80 kg
Mass of man 2, m₂ = 60 kg
One man pulls on the rope with a force of 250 N.
Let a₁ is acceleration of man 1,
F = m₁a₁

Let a₂ is acceleration of man 1,
F = m₂a₂

So, the acceleration of man 1 and 2 is
and
.
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:

Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ = 
A₂ = Area at the throat
A₂ = 
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or

or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s