Answer
Data provided:
mass of the block = 200 g = 0.2 Kg
Velocity at A = 0 m/s
Velocity at B = 8 m/s
distance of slide = 10 m
height of the block = 4 m
calculation for the block's potential energy
P = m g h
P = 0.2 x 9.8 x 4
P = 7.84 J
kinetic energy calculated as
Work done = P - KE
work = 7.84 - 6.14
work = 1.7 J
b) using the formula v² = u² + 2 a s
0 = 8² - 2 x a x 10
a = 3.2 m/s²
ma - μ mg = 0
Answer:
(a). The coefficient of permeability is .
(b). The seepage velocity is 0.0330 cm/s.
(c). The discharge velocity during the test is 0.0187 cm/s.
Explanation:
Given that,
Length = 175 mm
Diameter = 175 mm
Time = 90 sec
Volume= 405 cm³
We need to calculate the discharge
Using formula of discharge
Insert the values into the formula
(a). We will compute the coefficient of permeability
Applying the formula for permeability
Where, Q=discharge
l = length
A = cross-sectional area
h=constant head that causes flow
Plugging the value into the formula
(c). To ascertain the discharge velocity during the testing phase
Utilizing the discharge velocity formula
Substituting the values into the equation
The discharge velocity during the test measures 0.0187 cm/s.
Thus, (a). The coefficient of permeability is .
(b). The seepage velocity computes at 0.0330 cm/s.
(c). The observed discharge velocity during the test equals 0.0187 cm/s.
Answer:
The area that remains is 4.201 m²
Explanation:
Provided that
AB=D= 4 m (R=2 m)
The area of the half-circle AB is
A=2π
The area of the smaller half-circle is
a=5π/6 + 2√3/4 m²
a=5π/6 + √3/2 m²
<pThus, the remaining area = A - a= 2π - (5π/6 + √3/2) m²
The remaining area is expressed as 2π - (5π/6 + √3/2) m²
