Answer: Atomic number, Atomic Mass, Ionic radius
Explanation:
<span>Coefficient of static friction needs to be 1.1 or larger.
For this problem, we need to static friction to be at least as large as the centripetal acceleration that the car will experience. So let's get our formulas.
Centripetal acceleration:
F = mv^2/r
where
F = force
m = mass
v = velocity
r = radius of curve
Friction
F = mac
where
F = force
m = mass
a = gravitational acceleration
c = coefficient of friction
Since the frictional force has to be at least as large as the Centripetal force, let's set an inequality between them.
mv^2/r ≤ mac
v^2/r ≤ ac
v^2/(ar) ≤ c
Now let's convert km/h to a more convenient m/s.
104 km/h / 3600 s/h * 1000 m/km = 28.88888889 m/s
Let's substitute the known values into the inequality and calculate.
v^2/(ar) ≤ c
(28.88888889 m/s)^2/(9.8 m/s^2 * 78 m) ≤ c
834.5679012 m^2/s^2 / 764.4 m^2/s^2 ≤ c
1.091794743 ≤ c
Rounded to 2 significant figures gives a required coefficient of static friction of 1.1 or greater. This is a rather large value and indicates that the car is not at all likely to be capable of taking that curve at that speed. There are some things that can be done to mitigate the issue. Those being
1. Reduce the velocity.
2. Increase the normal force. Perhaps by aerodynamic means
3. Bank the curve.</span>
Answer:
The value of d is 183.51 m.
Explanation:
Given that,
Speed of car = 34.0 m/s
Suppose The car race in the circle parallel to the ground surface is at an angle 40°
The radius of circular path
Normal force acting on the car = N
We need to calculate the value of d
Using component of normal force
The horizontal component of normal force is equal to the gravitational force.
....(I)
The vertical component of normal force is equal to the centripetal force
.....(II)
Divided equation (I) by equation (II)
Put the value of g
Put the value into the formula
Hence, The value of d is 183.51 m.
Answer:
1066.67 m
Explanation:
Given:
v₀ = 96 km/h = 26.67 m/s
v = 48 km/h = 13.33 m/s
Δx = 800 m
Find: a
v² = v₀² + 2aΔx
(13.33 m/s)² = (26.67 m/s)² + 2a (800 m)
a = -0.333 m/s²
Given:
v₀ = 26.67 m/s
v = 0 m/s
a = -0.333 m/s²
Find: Δx
v² = v₀² + 2aΔx
(0 m/s)² = (26.67 m/s)² + 2 (-0.333 m/s²) Δx
Δx = 1066.67 m
Round as needed.
The equal velocity approach for duct size assumes that the air velocity in each duct segment is the same.
How fast is the air moving through a duct?
The most common unit of air velocity (distance traveled in a unit of time) is feet per minute (FPM). The amount of air passing past a location in the duct per period of time may be calculated by multiplying the airflow by the area of the duct. The standard unit for volume flow is cubic feet per minute (CFM).
What happens when the size of ducts changes to the airflow?
- Result for an image The equal velocity technique for duct size makes the assumption that air velocity is constant across the entire duct system.
- The main lesson to be learned from this is that when air goes from a bigger to a narrower duct, its velocity rises. The velocity drops when it transitions from a shorter to a bigger duct. The flow rate or the amount of air passing through the duct in cubic feet per minute is the same in all scenarios.
Learn more about air velocity here:
brainly.com/question/3255148
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