The velocity of a 1.3 kg remote-controlled car is plotted on the graph. The work of segment A is J.
2 answers:
Answer: 585 J
Explanation:
We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:
where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is
The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:
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Answer:
r= 98.3 mm
Explanation:
For rim
R= 0.209 m
M= 4.32 kg
For rods
m= 7.37 kg
L= 2 R= 2 x 0.209 = 0.418 m
The Total moment of inertia of the wagon
I=MR²+2 x 1/12 m L²
Now by putting the values
I=0.413 kg.m²
For disk:
t= 0.0462 m
Density ρ = 5990 kg/m³
Lets take r is the radius of disk
So the mass of the disc
m'=ρ πr² t
The moment of inertia of disc
I'=1/2 m'r²
I'=1/2 x r² x ρ πr² t
Given that
I = I'
1/2 x r² x ρ πr² t = 0.413 kg.m²
1/2 x r³ x ρ π t = 0.413
r³ x ρ π t = 0.826
r³=0.00095
r=0.0983 m
r= 98.3 mm