Relative to the ground, a car has a velocity of 15.3 m/s, directed due north. relative to this car, a truck has a velocity of 22.5 m/s, directed 52.0° north of east. what is the magnitude of the truck's velocity relative to the ground

The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

Explanation:

Given that,

Velocity of car relative to ground = 15.3 m/s

Velocity of truck relative to car = 22.5 m/s

We need to calculate the magnitude of the truck's velocity relative to the ground

We need to calculate the x component of the velocity

We need to calculate the y component of the velocity

Using Pythagorean theorem

Hence, The magnitude of the truck's velocity relative to the ground is 35.82 m/s.

    t = 0.719 s

Explanation:

For this fluid mechanics exercise, let's use Bernoulli's equation.

We are going to use index 1 for the liquid in the glass and index 2 for the liquid in the mouth

         P1 + ρ g y1 = P2 + ρ g y2

the speed of exit of the lemonade is small or equal in the two points.

The pressure in the glass is atmospheric pressure and the pressure in the mouth is p₂ - 0.75 atm

Atmospheric pressure  P_atm = 1 10⁵ P

       ρg (y₂-y₁) = P₁ - P₂

       Δy = (P₁ - P₂) / ρ g

       Δy = (1 - 0.75) 1 10⁵ / (1008 9.8)

       Δy = 2.53 m

This is the maximum length of the cigarette

to calculate the time we assume that the speed in the glass is very small, let's use Newton's second law

            F = m a

           P = F V

the volume is

           V = A Δy

            P A = (ρ A Δy)   a

            a = P /(ρ Δy)

            a = 0.25 1 105 / (1008 2.53)

             a = 9.8 m /s²

now with the accelerated we can use the kinematic relations to find the time

            y = y₁ t + ½ a t²

            v₁= 0

            y = √ 2y / a

            y =√ AR (2 2.53 / 9.8)

            t = 0.719 s