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lord [1]
2 years ago
13

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

Physics
1 answer:
LUCKY_DIMON [66]2 years ago
8 0
The mass of the object and its height in the gravitational field of the earth.
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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
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