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3 years ago

14

describe the motion of a pendulum in terms of kinetic and potential energy when it goes from its highest point to lowest point,

how would the energy change?

1 answer:

MatroZZZ [7]3 years ago

4 0

When it reaches it's peak, the energy is converted into potential as it slows down, then back to kinetic as it goes back to the lowest point.

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A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function

Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$$v(t)=\sin (\omega t) * \cos ^2(\omega t)$

To get the position equation we just need to integrate the above equation:

$$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t$

$$\mathrm{u}=\cos (\omega \mathrm{t})$

Then:

$$d u=-\sin (\omega t) d t$

$\Rightarrow d t=-d u / \sin (\omega t)$

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$$f(t)=\frac{\cos ^3(\omega t)}{3}+C$

To find the value of C, we use the fact that f(0) = 0.

$$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0$

C = -1 / 3

Then the position function is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

Integrating the velocity equation, we will see that the position equation is:

$$f(t)=\frac{\cos ^3(\omega t)-1}{3}$

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0

1 year ago

Infer the direction of the net force acting on a car as it slows down and turns right.

Kobotan [32]

Net force would be towards the right and back (opposite direction of motion) since it's slowing down (decelerating) and turning right.

6 0

3 years ago

Read 2 more answers

An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t

UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = $1.013 \times 10^{5}+ 100 \times 1000 \times 9.8$

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

$1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2$

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0

3 years ago

Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Andrew [12]

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

5 0

3 years ago

Plz help me with this activity :(

Marrrta [24]

As mass increases, the potential energy also increases, I hope that helped :)

8 0

3 years ago