Answer:
The car stops in 7.78s and does not spare the child.
Explanation:
In order to know if the car stops before the distance to the child, you take into account the following equation:
(1)
vo: initial speed of the car = 45km/h
a: deceleration of the car = 2 m/s^2
t: time
xo: initial distance to the child = 25m
x: final distance to the child = 0m
It is necessary that the solution of the equation (1) for time t are real.
You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:
You take the first value t1 because it has physical meaning.
The solution for t is real, then, the car stops in 7.78s and does not spare the child.
The formula for force exerted on/by a spring is
F = k*e where k is the spring constant and x is the distance stretched from
unstrained position. This should allow you to find what you need.
Using F = k x e,
where k is the spring constant,
and e is the extension,
The F is her weight = 45 X 0.80
= 36 N
Sitting = no movement
KE=0
Answer:
189 m/s
Explanation:
The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.
So, F = W
mv²/r = mg
v² = gr
v = √gr where v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m
So, v = √gr
v = √(9.8 m/s² × 3.63 × 10³ m)
v = √(35.574 × 10³ m²/s²)
v = √(3.5574 × 10⁴ m²/s²)
v = 1.89 × 10² m/s
v = 189 m/s
