Astudent conducts an experiment in which a cart is pulled by a variable applied force during a 2sec time interval. in trial 1, the student exerts the force on a cart of mass m. in trial 2, the student exerts the force on a cart of mass 3m. in trial 3, the student exerts the force on a cart of 5m. in which trial will the cart experience the greatest change in momentum from 0 sec to 2 sec?

change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

Explanation:

Let's look for the speed of the car

     F = m a

    a = F / m

We use kinematics to find lips

    v = v₀ + a t

    v = v₀ + (F / m) t

The moment is defined by

    p = m v

The moment change

    Δp = m v - m v₀

Let's replace the speeds in this equation

    Δp = m (v₀ + F / m t) - m v₀

    Δp = m v₀  + F t - m v₀  

    Δp = F t

We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures