Given,
the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
Answer:
33.33j+6.67i km/hr
Explanation:
From the law of conservation of momentum,
Applying,
mu+m'u' = V(m+m')............... Equation 1
Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.
Note: let j represent the north, and i represent the east
From the question,
Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i
Substitute these values into equation 1
1500*60j+1200*15i = V(1500+1200)
90000j+18000i = 2700V
V = (90000j+18000i)/2700
V = 33.33j+6.67i km/hr
Answer:
666 bc its on its way to my house
Explanation:
Answer:
3.44 metres
Explanation:
To determine the vector sum of the displacements Δd1 = 2.4 m [32° S of W]; Δd2 = 1.6 m [S]; and Δd3 = 4.9 m [27° S of E], resolve the given parameters into x - component and y - component.
Resolving into x - component
- 2.4cos32 + 4.9cos27 = 2.3306
Resolving into y - component
- 2.4sin32 - 4.9sin27 - 1.6 = - 2.553
The vector sum of the displacement will be
Sqrt( 2.3^2 + 2.6^2) =
Sqrt ( 11.81)
3.44 m
Therefore, the vector sum of the displacements is 3.44 metres