What variable(s) determine the amount of potential energy an object has?

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago

Unpolarized light of intensity I0 = 950 W/m2 is incident upon two polarizers. The first has its polarizing axis vertical, and th

Ket [755]

Answer:

Intensity of the light (first polarizer) (I₁) = 425 W/m²

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

Explanation:

Given:

Unpolarized light of intensity (I₀) = 950 W/m²

θ = 65°

Find:

a. Intensity of the light (first polarizer)

b. Intensity of the light (second polarizer)

Computation:

a. Intensity of the light (first polarizer)

Intensity of the light (first polarizer) (I₁) = I₀ / 2

Intensity of the light (first polarizer) (I₁) = 950 / 2

Intensity of the light (first polarizer) (I₁) = 425 W/m²

b. Intensity of the light (second polarizer)

Intensity of the light (second polarizer) (I₂) = (I₁)cos²θ

Intensity of the light (second polarizer) (I₂) = (425)(0.1786)

Intensity of the light (second polarizer) (I₂) = 75.905 W/m²

5 0

3 years ago

Something that can not be used up or depleted

Jet001 [13]

Answer:

oxygen

Explanation:

as it is in the air it can't be depleted or used up

5 0

3 years ago

Read 2 more answers

What is the difference in light that is refracted compared to the light that is reflected?

Kruka [31]

Answer:

Reflection is when light bounces off an object, while refraction is when light bends while passing through an object.

8 0

3 years ago

Read 2 more answers

A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. f(t) = 0.01t4 − 0.0

ivolga24 [154]

Since you solved a and b I will start with part c.
Part C
To answer this question we need to find zeros of a velocity function:
$v(t)=0.04t^3-0.06t^2$
We can factor this polynomial:
$v(t)=0.04t^3-0.06t^2=t^2(0.04t-0.06)$
Now it's pretty easy to find zeros. This function will be equal to zero when any of the factors are equal zero.
$t^2=0;\\ 0.04t-0.06=0$
We solve these two equations and we get our zeros:
$t_1=0; t_2=\frac{3}{2}$
The particle is at rest at t=0 and t=3/2.
Part D
To solve this we need to determine when our velocity function is greater than zero. We will use factored form.
We determine when each factor is greater than zero and with that information, we build the following table:
$\centering \label{my-label} \begin{tabular}{lllll} Range & -\infty & 0 & 3/2 & +\infty \\ t^2 & - & + & + & + \\ 0.04t-0.06 & - & - & + & + \\ t^2 (0.04t-0.06) & + & - & + & + \end{tabular}$
We can see, from the table, that our function is positive when $- \infty < t$ and t>3/2.
That is the range in which particle is moving in positive direction.
Part E
We know that distance traveled is given with:
$s(t)=0.01t^4 - 0.02t^3$
We simply plug in t=12 to find total distance traveled:
$s(12)=0.01(12)^4 - 0.02(12)^3=172.80 ft$
Part F
We know that acceleration is defined as a rate of change of velocity.
We find acceleration by taking the first derivative of velocity with respect to time.
$a(t)=\frac{dv}{dt}=(0.04t^3-0.06t^2)'=0.12t^2-0.12t$
To find acceleration after 1 second we simply plug in t=1s in above equation:
$a(1)=0.12-0.12=0$

7 0

3 years ago