Answer:
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Explanation:
For this exercise let's use the relationship between momentum and momentum.
I = F t = Δp
in this case the final velocity is zero
F t = 0 -m v₀
F = m v₀ / t
in order to answer the question we must assume that the two vehicles have the same mass and speed
concrete barrier
F₁ = -p₀ / 0.1
F₁ = - 10 p₀
barrier collapses
F₂ = -p₀ / 1
let's look for the relationship of the forces
F₁ / F₂ = 10
therefore the first out is 10 times greater than the second barrier
Answer:
ac = 3.92 m/s²
Explanation:
In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,
Frictional Force = Centripetal Force
where,
Frictional Force = μ(Normal Force) = μ(weight) = μmg
Centripetal Force = (m)(ac)
Therefore,
μmg = (m)(ac)
ac = μg
where,
ac = magnitude of centripetal acceleration of car = ?
μ = coefficient of friction of tires (kinetic) = 0.4
g = 9.8 m/s²
Therefore,
ac = (0.4)(9.8 m/s²)
<u>ac = 3.92 m/s²</u>
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So 
Answer: Because of the longitudinal motion of the air particles, there are regions in the air where the air particles are compressed together and other regions where the air particles are spread apart. These regions are known as compressions and rarefactions respectively
Explanation:
