Answer:
<em>1.01 W/m</em>
Explanation:
diameter of the pipe d = 30 mm = 0.03 m
radius of the pipe r = d/2 = 0.015 m
external air temperature Ta = 20 °C
temperature of pipe wall Tw = 150 °C
convection coefficient at outer tube surface h = 11 W/m^2-K
From the above,<em> we assumed that the pipe wall and the oil are in thermal equilibrium</em>.
area of the pipe per unit length A = 
= 
m^2/m
convectional heat loss Q = Ah(Tw - Ta)
Q = 7.069 x 10^-4 x 11 x (150 - 20)
Q = 7.069 x 10^-4 x 11 x 130 = <em>1.01 W/m</em>
Answer:
7.9060 m²
8.57 Volts
5.142×10⁻⁶ Joule
1.2×10⁻⁶ Coulomb
Explanation:
C = Capacitance between plates = 0.14 μF = 0.14×10⁻⁶ F
d = Distance between plates = 0.5 mm = 0.5×10⁻³ m
Q = Charge = 1.2 μC = 1.2×10⁻⁶ C
ε₀ = Permittivity = 8.854×10⁻¹² F/m
Capacitance
∴ Area of each plate is 7.9060 m²
Voltage
∴ Potential difference between the plates if the capacitor is charged to 1.2 μC is 8.57 Volts.
Energy stored
E=0.5CV²
⇒E = 0.5×0.14×10⁻⁶×8.57²
⇒E = 5.142×10⁻⁶ Joule
∴ Stored energy is 5.142×10⁻⁶ Joule
Charge
Q = CV
⇒Q = 0.14×10⁻⁶×8.57
⇒Q = 1.2×10⁻⁶ C
∴ Charge the capacitor carries before a spark occurs between the two plates is 1.2×10⁻⁶ Coulomb
Answer:
Explanation:
Given
charge on alpha particle=+2e
mass of alpha particle=
kg
Charge on gold nucleus=+79e
Velocity at r=1m is 
Using Energy conservation
Kinetic energy of particle will be converting to Potential energy as it approaches to nucleus
therefore
![\frac{1}{2}\left ( 6.64\times 10^{-27}\right )\left ( 1.9\times 10^{7}^2\right )=\frac{9\times 10^9\times 158\times \left ( 1.6\times 10^{-19}\right )}{y}\left [\frac{1}{r_0}-\frac{1}{1}\right ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%5Cleft%20%28%206.64%5Ctimes%2010%5E%7B-27%7D%5Cright%20%29%5Cleft%20%28%201.9%5Ctimes%2010%5E%7B7%7D%5E2%5Cright%20%29%3D%5Cfrac%7B9%5Ctimes%2010%5E9%5Ctimes%20158%5Ctimes%20%5Cleft%20%28%201.6%5Ctimes%2010%5E%7B-19%7D%5Cright%20%29%7D%7By%7D%5Cleft%20%5B%5Cfrac%7B1%7D%7Br_0%7D-%5Cfrac%7B1%7D%7B1%7D%5Cright%20%5D)
on solving we get
Answer:
(i) -556 rad/s²
(ii) 17900 revolutions
(iii) 11250 meters
(iv) -55.6 m/s²
(v) 18 seconds
Explanation:
(i) Angular acceleration is change in angular velocity over time.
α = (ω − ω₀) / t
α = (10000 − 15000) / 9
α ≈ -556 rad/s²
(ii) Constant acceleration equation:
θ = θ₀ + ω₀ t + ½ αt²
θ = 0 + (15000) (9) + ½ (-556) (9)²
θ = 112500 radians
θ ≈ 17900 revolutions
(iii) Linear displacement equals radius times angular displacement:
s = rθ
s = (0.100 m) (112500 radians)
s = 11250 meters
(iv) Linear acceleration equals radius times angular acceleration:
a = rα
a = (0.100 m) (-556 rad/s²)
a = -55.6 m/s²
(v) Angular acceleration is change in angular velocity over time.
α = (ω − ω₀) / t
-556 = (0 − 15000) / t
t = 27
t − 9 = 18 seconds
