Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²
2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads
3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²
4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J
5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²
6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²
7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s
8. s = Θ r = 266 rads * 0.33m = 88 m
Answer:
Explanation:
a ) starting from rest , so u = o and initial kinetic energy = 0 .
Let mass of the skier = m
Kinetic energy gained = potential energy lost
= mgh = mg l sinθ
= m x 9.8 x 70 x sin 30
= 343 m
Total kinetic energy at the base = 343 m + 0 = 343 m .
b )
In this case initial kinetic energy = 1/2 m v²
= .5 x m x 2.5²
= 3.125 m
Total kinetic energy at the base
= 3.125 m + 343 m
= 346.125 m
c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .
Answer:
16 J
Explanation:
It is given that,
Work done, W = 2 J
A spring is stretched by 2.0 cm from its equilibrium length
We need to find how much more work will be required to stretch it an additional 4.0 cm.
Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :
The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.
So, the required work done is 16 J.
Answer:
Kelly's weight would be 688.47 Newtons.
Explanation:
1 Kilogram would be 9.81 Newtons.
