Answer:
Explanation:
This problem is based on conservation of angular momentum.
moment of inertia of larger disc I₁ = 1/2 m r² , m is mass and r is radius of disc . I
I₁ = .5 x 20 x 5²
= 250 kgm²
moment of inertia of smaller disc I₂ = 1/2 m r² , m is mass and r is radius of disc . I
I₂ = .5 x 10 x 2.5²
= 31.25 kgm²
3500 rmp = 3500 / 60 rps
n = 58.33 rps
angular velocity of smaller disc ω₂ = 2πn
= 2π x 58.33
= 366.3124 rad /s
applying conservation of angular momentum
I₂ω₂ = ( I₁ +I₂) ω , ω is the common angular velocity
31.25 x 366.3124 = ( 250 +31.25) ω
ω = 40.7 rad / s .
A conjugate acid is formed from the base by accepting a proton from the acid .
A conjugate base is obtained from the Brownstead - Lowry acid when it looses a proton while the conjugate acid is obtained from the Brownstead - Lowry base when it accepts a proton. In the Brownstead - Lowry sense, acid base reaction involves the loss or gain of a proton.
Consider the hypothetical reaction; AH + :B ⇄ BH + :A. The specie BH is the conjugate acid while the specie :B is the Brownstead - Lowry base . The specie :A is the conjugate base while the specie AH is the Brownstead - Lowry acid.
Learn more about conjugate acid: brainly.com/question/10468518
Gravity acts to influence the vertical motion of the projectile, thus causing a vertical acceleration. The horizontal motion of the projectile is the result of the tendency of any object in motion to remain in motion at constant velocity.
Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:
Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m
The view factor between two coaxial parallel disks would be
Then the view factor between base to top surface of the cylinder becomes 
. From the summation rule
Then the net rate of radiation heat transfer from the disks to the environment is calculated as
Therefore the rate heat radiation is 780.76W
