Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) =
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
The student who did the most work is student 2 with 2500 Joules.
<u>Given the following data:</u>
To determine which of the students did the most work:
Mathematically, the work done by an object is given by the formula;
<u>For </u><u>student 1</u><u>:</u>
Work done = 600 Joules
<u>For </u><u>student 2</u><u>:</u>
Work done = 2500 Joules.
Therefore, the student who did the most work is student 2 with 2500 Joules.
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Force and Gravity, is what i think it is.