Answer:
F = -8440.12 N
the magnitude of the average force needed to hold onto the child is 8440.12 N
Explanation:
Given;
Mass of child m = 16 kg
Speed of each car v = 59.0 mi/h = 26.37536 m/s
Time t = 0.05s
Applying the impulse momentum equation;
Impulse = change in momentum
Ft = ∆(mv)
F = ∆(mv)/t
F = m(∆v)/t
Where;
F = force
t = time
m = mass
v = velocity
Since the final speed of the car is zero(at rest) then;
∆v = 0 - v = -26.37536 m/s
Substituting the given values;
F = 16×-26.37536/0.05
F = -8440.1152 N
F = -8440.12 N
the magnitude of the average force needed to hold onto the child is 8440.12 N
I would carry a 25 pound box up 5 flights, because it is lighter. You will wear out faster carrying a 50 pound box up 3 flights.
Answer: In a logical Pace forum subject to the distance
Explanation:
Answer:
Force on front axle = 6392.85 N
Force on rear axle = 8616.45 N
Explanation:
As we know that the weight of the car is balanced by the normal force on the front wheel and rear wheels
Now we know that
now we know that distance between the axis is 2.70 m and centre of mass is 1.15 m behind front axle
so we can write torque balance about its center of mass
now from above equation
now we have
now the other force is given as
