Airbags will deploy no matter from what angle your car is hit.
2 answers:
Answer:
<u>Airbags:</u>
" As the airbags comes in inbuilt form these days in most of the cars, as they fulfill the safety concerns of the passengers and make it sure that non of the individuals gets hurt in any accident faced during the travel."
<u>Angle of Deployment:</u>As the airbags will only deploy in the specific conditions provided, as the car must be speed over about the 25 km/hr and the accident must be more like a head-on-collision making the airbags to deploy. Or else if the car experiences an side wise crash with any other entity then it will never deploy in that case.
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Answer:
Explanation:
first write the newtons second law:
F=δma
Applying bernoulli,s equation as follows:
∑
Where, is the pressure change across the streamline and
is the fluid particle velocity
substitute for {tex]γ[/tex] and
for
integrating the above equation using limits 1 and 2.
there the bernoulli equation for this flow is
note: =density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular
Answer:
1) R₁ > R₃ > R₂ correct B , 2) the wire that dissipates the most is wire 2
Explanation:
1) The resistance of a wire is given by the expression
R =
where ρ is the resistivity of the material, l the length of the wire and A the area of the wire
The area is given by
A = π r² = π d² / 4
we substitute
R = (ρ 4 /π)
the amount in parentheses is constant for this case
let's analyze the situation presented, to find the resistance of each wire
* indicate l₁ = l₂ and d₂ = 2 d₁
the resistance of wire 1 is
R₁ = (ρ 4 /π)
the resistance of wire 2 is
R₂ = (ρ 4 /π) \frac{l_2}{d_2^2}
R₂ = (ρ 4 /π)
R₂ = (ρ 4 /π ) ¼
R₂ = ¼ R₁
* indicate that d₂ = d₃ and l₃ = 2 l₂
R2 = (ρ 4 /π)
the resistance of wire 3 is substituting the indicated condition
R3 = (ρ 4 /π 2) \frac{l_3}{d_3^2}
R3 = (ρ 4 /π)
R3 = 2 R₂
let's write the relations obtained
R₁ = (ρ 4 /π)
R₂ = ¼ R₁
R₃ = 2 R₂
let's write everything as a function of R1
R₁ =(ρ 4 /π)
R₂ = ¼ R₁
R₃ = ½ R₁
the resistance of the wire in decreasing order is
R₁ > R₃ > R₂
2) The power dissipated by a wire is
P = V I
the voltage is
V = I R
I = V / R
substituting
P = V² / R
therefore the power dissipated by each wire is
wire 1
P₁ = V² / R₁
wire 2
P₂ = V² / R₂
P₂ =
P₂ = 4 P₁
wire 3
P₃ = V² / R₃
P₃ =
P₃ = 2 P₁
Therefore, the wire that dissipates the most is wire 2