3 is B my dude and the reason nobody answered is because thousands of questions are asked every day. So not every question can be answered. It doesn't have to do with being lazy.
Answer:
1.25 kgm²/sec
Explanation:
Disk inertia, Jd =
Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²
Disk angular speed =
ωd = 0.1047 * 30 = 3.1416 rad/sec
Hollow cylinder inertia =
Jc = 3.7 * 0.40² = 0.592 kgm²
Initial Kinetic Energy of the disk
Ekd = 1/2 * Jd * ωd²
Ekd = 0.148 * 9.87
Ekd = 1.4607 joule
Ekd = (Jc + 1/2*Jd) * ω²
Final angular speed =
ω² = Ekd/(Jc+1/2*Jd)
ω² = 1.4607/(0.592+0.148)
ω² = 1.4607/0.74
ω² = 1.974
ω = √1.974
ω = 1.405 rad/sec
Final angular momentum =
L = (Jd+Jc) * ω
L = 0.888 * 1.405
L = 1.25 kgm²/sec
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, the formula to be used here is
ω = 2π/T
Where ω is the angular frequency (in rad/s)
T is the period - the time taken for Block A to complete one oscillation and return to it's original position.
To solve for this period T, the formula below should be used
T = 2π√m/k
where m is the mass of the object (Block A) and k is the spring constant (281 J/m²)
Answer:
a)= 0.025602u
b) = 23.848MeV
c) N = 1.546 × 10¹³
Explanation:
The reaction is
²₁H + ²₁H ⇄ ⁴₂H + Q
a) The mass difference is
Δm = 2m(²₁H) - m (⁴₂H)
= 2(2.014102u) - 4.002602u
= 0.025602u
b) Use the Einstein mass energy relation ship
The enegy release is the mass difference times 931.5MeV/U
E = (0.025602) (931.5)
= 23.848MeV
c)
the number of reaction need per seconds is
N = Q/E
= 59W/ 23.848MeV
N = 1.546 × 10¹³
Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
