Answer:
The car is going 0 km/h more than the bike
Explanation:
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
More force is needed for more mass. Therefore, if the mass is greater and the force is not enough then the object will less likely accelerate
Answer:
v = √2G 
/ R
Explanation:
For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)
Eo = K + U = ½ m1 v² - G m1 m2 / r1
Ef = - G m1 m2 / r2
When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf
Eo = Ef
½ m1v² - G m1 / R = - G m1 / R
v² = 2G (1 / R - 1 / Rinf)
If we do Rinf = infinity 1 / Rinf = 0
v = √2G / R
Ef = = - G m1 m2 / R
The mechanical energy is conserved
Em = -G m1 / R
Em = - G m1 / R
R = int ⇒ Em = 0
Answer:
a. dW = ∫pEsinθdθ b. W = p.E
Explanation:
a. We know torque τ = p × E = pEsinθ where θ is the angle between p and E
Let the torque τ rotate the dipole by an amount dθ. So, the workdone dW = ∫τdθ = ∫pEsinθdθ
b. So, the total work done is gotten by integrating from 90 to θ. So,
W = ∫₉₀⁰dW
= ∫₉₀⁰pEsinθdθ
= pE∫₉₀⁰sinθdθ
= pE(cosθ - cos90)
=pEcosθ
= p.E
