The answer would be center of mass, B
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:
The energy returns to the weightlifter's muscles, where it is dissipated as heat.
Explanation:
The energy returns to the weightlifter's muscles, where it is dissipated as heat. As long as the weightlifter controls the weight's descent, their muscles are acting as an overdamped shock absorber, as if the weight were sitting on a piston containing very thick fluid, slowly compressing it downward (and slightly heating up the fluid in the process). Since muscles are complicated biological systems and not simple pistons, they require metabolic energy to maintain tension throughout the controlled descent, so the weightlifter feels like they're putting energy into the weight, even though the weight's gravitational potential energy is being converted into heat within the lifter's muscles.
The answer is A
Good luck!
Answer:
a) E = 0
b) 
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) 
From which we have;
E = 0/A = 0
E = 0
b)
By Gauss theorem, we have;
Therefore, we get;
The electrical field outside the spherical shell
Therefore, we have;
