The answer of this question is B. 22x + 20
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Answer:
17304 J
Explanation:
Complete statement of the question is :
In the winter activity of tubing, riders slide down snow covered slopes while sitting on large inflated rubber tubes. To get to the top of the slope, a rider and his tube, with a total mass of 84 kg , are pulled at a constant speed by a tow rope that maintains a constant tension of 350 N .
Part A
How much thermal energy is created in the slope and the tube during the ascent of a 30-m-high, 120-m-long slope?
Solution :
= tension force in the tow rope = 350 N
= length of the incline surface = 120 m
= work done by tension force = ?
The tension force acts parallel to incline surface, hence work done by tension force is given as
= height gained by the rider = 30 m
= total mass of rider and tube = 84 kg
Potential energy gained is given as
= Thermal energy created
Using conservation of energy
(50 gal / 5 min) x (.0037854 m³/gal) x (1 min / 60 sec)
= (50 · 0.0037854 · 1) / (5 · 60) m³/sec
= 0.000631 m³/sec
A seesaw remains stationary when two students of equal weight sit on the ends
c
