Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251
Answer: pretty sure 14 pounds
Explanation:
Answer:
A. Linkedln
Explanation:
In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:
A. Linkedln
All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.
Question is not complete and the missing part is;
A coin of mass 0.0050 kg is placed on a horizontal disk at a distance of 0.14 m from the center. The disk rotates at a constant rate in a counterclockwise direction. The coin does not slip, and the time it takes for the coin to make a complete revolution is 1.5 s.
Answer:
0.828 m/s
Explanation:
Resolving vertically, we have;
Fn and Fg act vertically. Thus,
Fn - Fg = 0 - - - - eq(1)
Resolving horizontally, we have;
Ff = ma - - - - eq(2)
Now, Fn and Fg are both mg and both will cancel out in eq 1.
Leaving us with eq 2.
So, Ff = ma
Now, Frictional force: Ff = μmg where μ is coefficient of friction.
Also, a = v²/r
Where v is linear speed or velocity
Thus,
μmg = mv²/r
m will cancel out,
Thus, μg = v²/r
Making v the subject;
rμg = v²
v = √rμg
Plugging in the relevant values,
v = √0.14 x 0.5 x 9.8
v = √0.686
v = 0.828 m/s
