Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
Answer:
New Resistance = 0.5556 ohm
Explanation:
Resistance = resistivity * length /area
Here since resistivity and length are constant, we only need to see how the resistance increases or decreases with change in area.
New Area = pi * (3*D)^2 / 4
Old Area = pi * D^2 / 4
The ratio of new area / old area is :
Since area increases 9 times, and it is inversely proportional to resistance:
Resistance decreases by 9 times.
So, old resistance = Voltage / Current = 10 / 2 = 5 ohm
New Resistance = 5 / 9 = 0.5556 ohm (decreases by 9 times)
Answer:
Explanation:
PE = mgh = 60(9.8)(2.0) = 1176 J
Answer:
The magnitude of the force is 34.59 N.
Explanation:
Given that,
Inside pressure 
Area 
Outside pressure = 1 atm
We need to calculate the magnitude of the force
Using formula of force
Where, 
=inside Pressure
=outside Pressure
A = area
Put the value into the formula
Hence, The magnitude of the force is 34.59 N.
