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 .
Spinning a marshmallow over a fire is effective maybe if you hang it over the fire and heat it up equally on each side
Answer:
Vf=3
Explanation:
you must first write your data
data before impact
M1=1000 M2=5000
V1=0 m/s V2 =0m/s
data after impact
M1=1000 M2=5000
V1=15m/s V2=?
M1V1 +M2V2=M1V1 +M2V2f
(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf
0=15000+5000Vf
- 15000÷5000=5000Vf÷5000
Vf= -3
Vf =3
Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb
