Answer:
i have no clue i just need brailnly points
Explanation:
I think the word would be increases. I don't know what the options are though.
Answer:
a) ω = 91.1 rad/s
b) KEr = 15.0 J
Explanation:
I = (2/3)mR²
ω = v/R
PE = KE
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(2/3)mR²(v/R)²
2gh = v² + (2/3)v²
2gh = 5/3v²
v² = 6gh/5
ω² = v²/R² = 6gh/5R²
a. At what rate was it rotating at the base of the hill?
ω = √(6(9.81)(9.00) / (5(0.226/2)²) = 91.1 rad/s
b. How much rotational kinetic energy did it then have?
KE = ½Iω² = ½(⅔)(0.426)(0.113)²(91.1²) = 15.044616
Elastic collision is when kinetic energy before = kinetic energy after
Ek= 1/2mv^2
total before
Ek=1/2(2)(2.2^2) = 4.84 J
total after
Ek= 1/2(2+4)(v^2) = 3v^2
Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s
Answer:
Fint = - 67.094 N
Explanation:
Given
W = 71.4 N
∅ = 20º
Fint = ?
The interaction partner of the Normal Force has equal magnitude and opposite direction to the Normal Force, and it is a contact force exerted by the crate on the ramp, because the interaction parteners are the same force acts on the two object.
Then: Fint = - N = - Wy' = - W*Cos ∅
⇒ Fint = - 71.4 N*Cos 20º = - 67.094 N