Physics, 02.07.2019 02:00 m10479

Two identical masses, a and b, slide down frictionless ramps, both starting at a height of 3m above the ground. object a slides down a very steep ramp, while b slides down a very gradual incline. which object is travelling faster at the bottom of the ramp? a. object a b. object b c. neither, they have the same speed

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Answer from: irisprats

c. Neither, they have the same speed.

In both the cases, there will be work done by the gravitational force. Since, the gravitational force is a conservative force, its work done does not depend on the path as it only depends on the initial and final position of the objects. Since, the initial and final positions are same for both the objects, the work done by gravitational force will be same in both the cases.

Hence, Work done by gravitational force = Change in the kinetic energy

So, the kinetic energy will be same for both the object.

Since, the masses are same: Kinetic energy = $\frac{1}{2}mv^2$

Hence, the speed will also be same for both the objects.

Correct option: C

Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ 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 $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

Answer from: Quest

Answerthe energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.explanationwe define electric potential as the ratio electric potential energy to the charge, $v = \dfrac{e_p}{q}$ then the following gives us the potential difference, $\delta v = v_b - v_a$ $\begin{aligned} \delta v & = \dfrac{e_p_b}{q} - \dfrac{e_p_a}{q} \\ & = \dfrac{\delta e_p}{q} \end{aligned}$ the formula $\delta v = \delta e_p / q$ may be on your formula sheet.the change in electric potential energy would be the energy that would needed to move it from point a to point b. solving for that, we get $\delta e_p = q\delta v$ using q = 8.00 × 10⁻¹⁹ c and δv = 2.00 v, we get $\begin{aligned} \delta e_p & = (8.00\times10^{-19} \text{ c})(2.00\text{ v}) \\ & = 1.60\times 10^{-18}\text{ j} \end{aligned}$ the energy required is 1.60 × 10⁻¹⁸ j, to three significant figures.

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Two identical masses, a and b, slide down frictionless ramps, both starting at a height of 3m above...

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