Answer: The drag force goes up by a factor of 4
Explanation:
The <u>Drag Force</u> equation is:
(1)
Where:
is the Drag Force
is the Drag coefficient, which depends on the material
is the density of the fluid where the bicycle is moving (<u>air in this case)
</u>
is the transversal area of the body or object
the bicycle's velocity
Now, if we assume , and do not change, we can rewrite (1) as:
(2)
Where 
groups all these coefficients.
So, if we have a new velocity 
, which is the double of the former velocity:
(3)
Equation (2) is written as:
(4)
Comparing (2) and (4) we can conclude<u> the Drag force is four times greater when the speed is doubled.</u>
true or false: a image created by an object located inside the focal point is virtual, enlarged, and upright
True :D
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
Answer:
(C) greater than zero but less than 45° above the horizontal
Explanation:
The range of a projectile is given by R = v²sin2θ/g.
For maximum range, sin2θ = 1 ⇒ 2θ = sin⁻¹(1) = 90°
2θ = 90°
θ = 90°/2 = 45°
So the maximum horizontal distance R is in the range 0 < θ < 45°, if θ is the angle above the horizontal.
