Answer:
2.06 m³/s
Explanation:
diameter of pipe, d = 0.81 m
diameter of constriction, d' = 0.486 m
radius, r = 0.405 m
r' = 0.243 m
density of oil, ρ = 821 kg/m³
Pressure in the pipe, P = 7970 N/m²
Pressure at the constriction, P' = 5977.5 N/m²
Let v and v' is the velocity of fluid in the pipe and at the constriction.
By use of the equation of continuity
A x v = A' x v'
r² x v = r'² x v'
0.405 x 0.405 x v = 0.243 x 0.243 x v'
v = 0.36 v' .... (1)
Use of Bernoulli's theorem
7970 + 0.5 x 821 x 0.36 x 0.36 x v'² = 5977.5 + 0.5 x 821 x v'² from (i)
1992.5 = 357.3 v'²
v' = 5.58 m/s
v = 0.36 x 5.58
v = 2 m/s
Rate of flow = A x v = 3.14 x 0.405 x 0.405 x 2 x 2 = 2.06 m³/s
Thus the rate of flow of volume is 2.06 m³/s.
Answer:
a) Focal length of the lens is 8 cm which is a convex lens
b) 6 cm
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Explanation:
u = Object distance = 4 cm
v = Image distance = -8 cm
f = Focal length
Lens Equation
a) Focal length of the lens is 8 cm which is a convex lens
Magnification
b) Height of image is 2×3 = 6 cm
Since magnification is positive the image upright
c) The lens is a convex lens and produces a virtual image which is upright and two times larger than the object.
Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
(1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:
Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:
The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
(2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
With M the mass of the sphere an R its radius, then:
Then (2) is:
Answer and Explanation:
A. We have temperature t = 32
Speed of sound, s = 1087.5
As t increases by 1⁰f speed increases by 1.2
So that
S = 1088.6
T= 33⁰f
We have 2 equations
1087.5 = k(32) + c
1088.6 = k(33) + c
Subtracting both equations
(33-32)k = 1088.6-1087.5
K = 1.1
b.). S = kT + c
1087.5 = 32(1.1) + c
Such that
C = 1052.3
Therefore
S = 1.1(t) + 1052.3
C.). S = 1.1t + 1052.3
We make t subject of the formula
T = s/1.1 - 1052.3/1.1
T = 0.90(s) - 956.3
D. This means that We have temperature to rise by 0.90 whenever speed is increased
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 × 
s
speed = 0.8 c = 0.8 × 3 × 
to find out
expected distance over
solution
we know c is speed of light in air is 3 × m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 × × 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
