Question
Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s. What is its angular momentum? Express your answer in kilogram meters squared per second.
Answer:

Explanation:
The angular momentum L of the baton moving about an axis perpendicular to it, passing through the center of the baton is,

Here, l is the length of the baton.
Substitute 0.120 kg for m, 3 rads/s for ![\omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}](https://tex.z-dn.net/?f=%5Comega%5B%5Ctex%5D%20and%200.8%20m%20for%20l%20%5Btex%5D%5Cbegin%7Barray%7D%7Bc%7D%5C%5CL%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7Dm%7Bl%5E2%7D%5Comega%20%5C%5C%5C%5C%20%3D%20%5Cfrac%7B1%7D%7B%7B12%7D%7D%5Cleft%28%20%7B0.120%7B%5Crm%7B%20kg%7D%7D%7D%20%5Cright%29%7B%5Cleft%28%20%7B%7B%5Crm%7B80%7D%7D%7B%5Crm%7B.0%20cm%7D%7D%7D%20%5Cright%29%5E2%7D%7B%5Cleft%28%20%7B%5Cfrac%7B%7B1%20%5Ctimes%20%7B%7B10%7D%5E%7B%20-%202%7D%7D%7B%5Crm%7Bm%7D%7D%7D%7D%7B%7B1%7B%5Crm%7B%20cm%7D%7D%7D%7D%7D%20%5Cright%29%5E2%7D%5Cleft%28%20%7B%7B%5Crm%7B3%7D%7D%7B%5Crm%7B.00%20rad%2Fs%7D%7D%7D%20%5Cright%29%5C%5C%5C%5C%20%3D%200.0192%7B%5Crm%7B%20kg%7D%7D%20%5Ccdot%20%7B%7B%5Crm%7Bm%7D%7D%5E%7B%5Crm%7B2%7D%7D%7D%7B%5Crm%7B%2Fs%7D%7D%5C%5C%5Cend%7Barray%7D)
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units
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:
Can you specify what is the question asking for