Answer:
Explanation:
a )
Depth of hole from surface of water d = .50 m - .03 m = .47 m
velocity of efflux v = √ 2gd
v = √ (2 x 9.8 x .47 )
v = 3.03 m /s
b )
Volume flow rate = π R² v where R is radius of hole at the bottom .
= 3.14 x ( .005 ) ² x 3.03 m/s
= 2.378 x 10⁻⁴ m³ /s
c )
Volume of water collected in 60 s
= 2.378 x 10⁻⁴ x 60
= 1.4268 x 10⁻² m³
If height attained in collecting container be h
π R² h = 1.4268 x 10⁻² m³ where R is radius of container
3.14 x ( .1 )² x h = 1.4268 x 10⁻²
h = .4544 m .
Pressure at the bottom of container = hρ g
where h is height of water , ρ is density of water
Pressure = .4544 x 1000 x 9.8 N /m²
= 4453.12 N /m²
Answer:
1.022 x 103 N.m
Explanation:
Solution
Given:
The weight of the block of mass m₂ is :
w₂ = m₂*g
Where
w₂ = 39 x 9.8 = 382.2 N
Then,
The weight of the block of mass m₁
w₁= m₁*g;
so,
w₁ = 12 x 9.8 = 117.5 N
Thus,
The tension wrapped in cord on drum (80 cm) T₁ = F - w₁
Now,
T₁ = 1200 - 117.5
T₁ = 1082.5 N
The tension wrapped in the cord on drum (41 cm) T₂ = w₂;
T₂ = 382.2 N
Hence,
We calculate net torque on the center of the drum:
The net torque = T₁ x 0.8 + T₂ x 0.41;
= 1082.5 x 0.8 + 382.2 x 0.41;
= 1.022 x 103 N.m
Therefore, the resulting torque applied to the system is 1.022 x 103 N.m
The answer is C.
The magnitude is greater and the motion is going in the direction of the arrow meaning it’s a straight line in the direction will not change