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:
here given is a weight
then force becomes mg
that is F=Mg
=4*9.8
then by using the formula
F=Ma
a=F/M
=4*9.8/9.8
=4
Explanation:
Frequency= speed/ wavelength
=0.5m/s divided 0.1.m
=5.0 Hz
The answer would be letter D.
Answer:
Explanation:
Let the volume of the unknown bulb = X L
The volume of the system , after opening valve = (X + 0.72 L )
Use Boyles law gas equation,
P1V1 = P2V2 ( at temperature is constant )
Given:
P1 = 1.2 atm
P2 = 683 torr
Converting mmHg to atm,
1 atm = 760 mmHg(torr)
683 mmHg = 683/760
= 0.8987 atm
1.2X = 0.8987*(X + 0.720)
1.2X = 0.8987X + 0.6471
0.3013X = 0.6471
X = 2.15 L
Answer:
W = 47040 J
Explanation:
Given that,
The mass of a student, m = 60 kg
Height of the tower, h = 80 m
We need to find the work done in climbing the tower. The work done is given by :
W = mgh
So,
W = 60 × 9.8 × 80
W = 47040 J
So, the required work done is 47040 J.
