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:
I didn't understand that all please tell me in Russia
Explanation:
This is simple.
Convert the speed of 45km/h to metres per second (m/s):
45 * 1000m = 45000m per hour.
1 hour = 60 seconds * 60 minutes = 3600 seconds.
45000m/h / 3600 = 12.5m/s
(A quicker way is just to divide by 3.6)
So in 20 seconds it would cover:
12.5m/s * 20s = 250m.
A car travelling at 45km/h would travel 250 metres in 20s.
Answer:
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
Explanation:
For this exercise we are asked to change the length of the bar by the action of a force applied along its length, in this case we focus on the expression of longitudinal elasticity
F / A = Y ΔL/L
where F / A is the force per unit length, ΔL / L is the fraction of the change in length, and Y is Young's modulus.
In this case the bars are made of the same material by which Young's modulus is the same for all
ΔL / L = (F / A) / Y
the area of the bar is the area of a circle
A = π r² = π d² / 4
A = π / 4 d²
we substitute
ΔL / L = (F / Y) 4 /πd²
changing length
ΔL = (F / Y 4 /π) L / d²
The amount between paracentesis are all constant in this exercise, let's look for the longitudinal change
a) values given d and 3L
ΔL = cte 3L / d²
ΔL = cte L /d² 3
To find the percentage, we must divide the change in magnitude by its value and multiply by 100.
ΔL/L % = [(F /Y 4/π 1/d²) 3L ] / 3L 100
ΔL/L % = cte 100%
b) 3d and L value, we repeat the same process as in part a
ΔL = cte L / 9d²
ΔL = cte L / d² 1/9
ΔL / L% = cte 100/9
ΔL / L% = cte 11%
c) 2d and 2L value
ΔL = (cte L / d ½
)/ 2L
ΔL/L% = cte 100/4
ΔL/L% = cte 25%
d) value 4d and L
ΔL = cte L / d² 1/16
ΔL/L % = cte 100/16
ΔL/L % = cte 6.25%
The highest percentage of change corresponds to the thinnest rod, the correct answer is a
