We know, W = k. q₁q₂/ [1/r₁² - r₂² ]
Here, r₁ = R
r₂ = ∞
Substitute their values,
W = k. q₁q₂/ [1/R - 1/∞]
W = k. q₁q₂/ [1/R]
W = k. q₁q₂/ R
Hope this helps!
Answer:
Part a)
Moment of inertia of the cylinder is given as
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as
Explanation:
As we know that the inclined plane is of length L = 3 m
and its inclination is given as 25 degree
so we know that acceleration of center of mass of the cylinder is constant so we will have
so we have
now we know that
Now we have know that final speed of the cylinder due to pure rolling is given as
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as
Answer:
Nails rusting over time as they are exposed to oxygen
Explanation:
New substance with new property is formed
It have also changed the chemical property of the substance
It is difficult to reverse the change etc...
<h2>
Answer:</h2>
<em>The balloon will attach to the wall because the balloon's negative charges will drive the electrons in the wall to shift to the other side of their atoms, leaving the wall's surface positively charged.</em>
Explanation:
Elongation of the wire is:
ΔL = F L₀ / (E A)
where F is the force,
L₀ is the initial length,
E is Young's modulus,
and A is the cross sectional area.
ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)
ΔL = T (1.25×10⁻⁶ m/N)
T = (80,000 N/m) ΔL
Draw a free body diagram of the mass at the bottom of the circle. There are two forces: tension force T pulling up and weight force mg pulling down.
Sum of forces in the centripetal direction:
∑F = ma
T − mg = mv²/r
T − mg = mω²r
T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)
T − 147 N = (2368.7 N/m) (0.5 m + ΔL)
Substitute:
(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)
(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL
(797631.3 N/m) ΔL = 1331.35 N
ΔL = 0.00167 m
ΔL = 1.67 mm