Answer:
a) w₂ = 5.16 rev / s
, b) K₀ = 29.55 J, K₂ = 190.88 J
Explanation:
The skater forms an isolated system whereby its angular momentum is preserved
Initial. With outstretched arms
L₀ = I₀ w₀
Final. With arms stuck
L₂ = I₂W₂
L₀ = L₂
I₀ w₀ = I₂ w₂
w₂ = I₀ / I₂ w₀
Let's calculate
w₂ = 0.800 2.34 / 0.363
w₂ = 5.16 rev / s
b) The kinetic energy is
K = ½ I w²
Let's reduce to the SI system
w₀ = 0.800 rev / s (2π rad / 1 rev) = 5.026 rad / s
w₂ = 5.16 rev / s (2π rad / 1 rev) = 32.42 rad / s
The initial kinetic energy
K₀ = ½ I₀ w₀²
K₀ = ½ 2.34 5,026²
K₀ = 29.55 J
The final kinetic energy
K₂ = ½ I₂ w₂²
K₂ = ½ 0.363 32.42²
K₂ = 190.88 J
When it rains, the growth aids in the soil drift
Answer:
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Explanation:
You can find the field using Gauss's Law:
the surface S is an "infinite long" cylinderr of radio r.
E(r)=
λ=-92.0 μC/m, ε=8.85.10^-12
E(0.1m)=-16.53.10^6 V.m
E(0.465m)=-3.55.10^6 V.m
E(1.3m)=-1.27^6 V.m
Answer:
C. Momentum is conserved but not kinetic energy.
Explanation:
This case represents an entirely inelastic collision, that is, a collision between the car and the truck that reduces total kinetic energy of the entire system, whereas linear momentum is conserved. Hence, correct answer is C.
Answer:
0.011 m.
Explanation:
Energy stored in the spring = Energy of the projectile.
1/2ke² = mgh ................ Equation 1
Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.
make e the subject of the equation
e = √(2mgh/k)............................. Equation 2
Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m
Constant: g = 9.8 m/s²
Substitute into equation 2
e = √(2×0.015×5/1200)
e = √(0.15/1200)
e = √(0.000125)
e = 0.011 m.