Answer:
I = 113.014 kg.m^2
m = 2075.56 kg
wf = 3.942 rad/s
Explanation:
Given:
- The constant Force applied F = 300 N
- The radius of the wheel r = 0.33 m
- The angular acceleration α = 0.876 rad / s^2
Find:
(a) What is the moment of inertia of the wheel (in kg · m2)?
(b) What is the mass (in kg) of the wheel?
(c) The wheel starts from rest and the tangential force remains constant over a time period of t= 4.50 s. What is the angular speed (in rad/s) of the wheel at the end of this time period?
Solution:
- We will apply Newton's second law for the rotational motion of the disc given by:
F*r = I*α
Where, I: The moment of inertia of the cylindrical wheel.
I = F*r / α
I = 300*0.33 / 0.876
I = 113.014 kg.m^2
- Assuming the cylindrical wheel as cylindrical disc with moment inertia given as:
I = 0.5*m*r^2
m = 2*I / r^2
Where, m is the mass of the wheel in kg.
m = 2*113.014 / 0.33^2
m = 2075.56 kg
- The initial angular velocity wi = 0, after time t sec the final angular speed wf can be determined by rotational kinematics equation 1:
wf = wi + α*t
wf = 0 + 0.876*(4.5)
wf = 3.942 rad/s
Answer:
3.62 V
Explanation:
L = 80 cm = 0.8 m
f = 15 rps
B = 60 m T = 0.060 T
ω = 2 x π x f = 2 x 3.14 x 15 = 94.2 rad/s
v = r ω
here, r be the radius of circular path. Here r = length of rod = L
v = 0.80 x 94.2 = 75.36 m/s
The motional emf is given by
e = B v L = 0.060 x 75.36 x 0.8 = 3.62 V
#1
Volume of lead = 100 cm^3
density of lead = 11.34 g/cm^3
mass of the lead piece = density * volume
so its weight in air will be given as
now the buoyant force on the lead is given by
now as we know that
so by solving it we got
V = 11.22 cm^3
(ii) this volume of water will weigh same as the buoyant force so it is 0.11 N
(iii) Buoyant force = 0.11 N
(iv)since the density of lead block is more than density of water so it will sink inside the water
#2
buoyant force on the lead block is balancing the weight of it
(ii) So this volume of mercury will weigh same as buoyant force and since block is floating here inside mercury so it is same as its weight = 11.11 N
(iii) Buoyant force = 11.11 N
(iv) since the density of lead is less than the density of mercury so it will float inside mercury
#3
Yes, if object density is less than the density of liquid then it will float otherwise it will sink inside the liquid
We need the power law for the change in potential energy (due to the Coulomb force) in bringing a charge q from infinity to distance r from charge Q. We are only interested in the ratio U₁/U₂, so I'm not going to bother with constants (like the permittivity of space).
<span>The potential energy of charge q is proportional to </span>
<span>∫[s=r to ∞] qQs⁻²ds = -qQs⁻¹|[s=r to ∞] = qQr⁻¹, </span>
<span>so if r₂ = 3r₁ and q₂ = q₁/4, then </span>
<span>U₁/U₂ = q₁Qr₂/(r₁q₂Q) = (q₁/q₂)(r₂/r₁) </span>
<span>= 4•3 = 12.</span>
