Consider a linear harmonic oscillator and let, yo and y, be its real, normalized ground and first

How does the movement of particles of matter change when temperature decreases

Zanzabum

INCREASE in temperature of the material practically increase the energy of the particles. which increases their motion due to increase in energy . thus when the temperature is decreased the energy level decreases which causes the particle's motion to slow down.. the motion of the particle is highly reduced when the temperature is lowered

4 0

4 years ago

If the lily pads are spaced 2.4 m apart and Ferdinand jumps with a speed of 5 m/s taking 0.6 s to go from lily pad to lily pad,

Zina [86]

Answer:

$36.87^{\circ}$

Explanation:

v = Velocity of Ferdinand = 5 m/s

$\theta$ = Angle of jump

T = Time taken = 0.6 s

R = Distance between lily pads = 2.4 m

Horizontal range is given by

$R=v_xT\\\Rightarrow R=vcos\theta T\\\Rightarrow cos\theta=\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{R}{vT}\\\Rightarrow \theta=cos^{-1}\dfrac{2.4}{5\times 0.6}\\\Rightarrow \theta=36.87^{\circ}$

The angle at which Ferdinand make each of his jumps is $36.87^{\circ}$

8 0

4 years ago

POR FAVOR AYUDENME A COMLETARLO

Zigmanuir [339]

Answer:

English please?

Explanation:

8 0

3 years ago

Which of the following statements are true? Select all correct responses. Choose all that apply. Choose all that apply. The tota

Ne4ueva [31]

<h3><u>Answer</u>;</h3>

-The total momentum of an isolated system is constant.

-The total momentum of any number of particles is equal to the vector sum of the momenta of the individual particles.

-The vector sum of forces acting on a particle equals the rate of change of momentum of the particle with respect to time.

<h3><u>Explanation</u>;</h3>

Momentum is a vector quantity, and therefore we need to use vector addition when summing together the momenta of the multiple bodies which make up a system.
The vector sum of forces acting on a particle is equivalent to the rate of change of momentum of the particle with respect to time. This is according to the Newton's second Law of motion. In mathematical terms, ֿF = d ֿp/dt, that is F= ma.
According to the Law of conservation of Momentum, or a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

3 0

3 years ago

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.?a.What is the speed of the mountain lio

Arisa [49]

Answer:

What is the speed of the mountain lion as it leaves the ground?

9.98m/s

At what angle does it leave the ground?

50.16°

Explanation:

This is going to be long, so if you want to see how it was solved refer to the attached solution. If you want to know the step by step process, read on.

To solve this, you will need use two kinematic equations and SOHCAHTOA:

$d = v_it + \dfrac{1}{2}at^{2}\\\\vf = vi + at$

With these formulas, we can derive formulas for everything you need:

Things you need to remember:

A projectile at an angle has a x-component (horizontal movement) and y-component (vertical movement), which is the reason why it creates an angle.

Treat them separately.

At maximum height, the vertical final velocity is always 0 m/s going up. And initial vertical velocity is 0 m/s going down.

Horizontal movement is not influenced by gravity.
acceleration due to gravity (a) on Earth is constant at 9.8m/s

First we need to take your given:

10.0 m long (horizontal) and maximum height of 3.0m (vertical).

$d_x=10.0m\\d_y=3.0m$

What your problem is looking for is the initial velocity and the angle it left the ground.

Vi = ? Θ =?

Vi here is the diagonal movement and do solve this, we need both the horizontal velocity and the vertical velocity.

Let's deal with the vertical components first:

We can use the second kinematic equation given to solve for the vertical initial velocity but we are missing time. So we use the first kinematic equation to derive a formula for time.

$d_y=V_i_yt+\dfrac{1}{2}at^{2}$

Since it is at maximum height at this point, we can assume that the lion is already making its way down so the initial vertical velocity would be 0 m/s. So we can reduce the formula:

$d_y=0+\dfrac{1}{2}at^{2}$

$d_y=\dfrac{1}{2}at^{2}$

From here we can derive the formula of time:

$t=\sqrt{\dfrac{2d_y}{a}}$

Now we just plug in what we know:

$t=\sqrt{\dfrac{(2)(3.0m}{9.8m/s^2}}\\t=0.782s$

Now that we know the time it takes to get from the highest point to the ground. The time going up is equal to the time going down, so we can use this time to solve for the intial scenario of going up.

$vf_y=vi_y+at$

Remember that going up the vertical final velocity is 0m/s, and remember that gravity is always moving downwards so it is negative.

$0m/s=vi_y+-9.8m/s^{2}(0.782s)\\-vi_y=-9.8m/s^{2}(0.782s)\\-vi_y=-7.66m/s\\vi_y=7.66m/s$

So we have our first initial vertical velocity:

Viy = 7.66m/s

Next we solve for the horizontal velocity. We use the same kinematic formula but replace it with x components. Remember that gravity has no influence horizontally so a = 0:

$d_x=V_i_xt+\dfrac{1}{2}0m/s^{2}(t^{2})\\d_x=V_i_xt$

But horizontally, it considers the time of flight, from the time it was released and the time it hits the ground. Also, like mentioned earlier the time going up is the same as going down, so if we combine them the total time in flight will be twice the time.

T= 2t

T = 2 (0.782s)

<em>So we use this in our formula:</em>

<em> $d_x=V_i_xT\\\\10.0m=Vi_x(1.564s)\\\\\dfrac{10.0m}{1.564s}=V_i_x\\\\6.39m/s=V_i_x$ </em>

Vix=6.39m/s

Now we have the horizontal and the vertical component, we can solve for the diagonal initial velocity, or the velocity the mountain lion leapt and the angle, by creating a right triangles, using vectors (see attached)

To get the diagonal, you just use the Pythagorean theorem:

c²=a²+b²

Using it in the context of our problem:

$Vi^{2}=Viy^2+Vix^2\\Vi^2=(7.66m/s)^2+(6.39m/s)^2\\\sqrt{Vi}=\sqrt{(7.66m/s)^2+(6.39m/s)^2}\\\\Vi=9.98m/s$

The lion leapt at 9.98m/s

Using SOHCAHTOA, we know that we can TOA to solve for the angle, because we have the opposite and adjacent side:

$Tan\theta=\dfrac{O}{A}\\\\Tan\theta=\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{V_i_y}{V_i_x}\\\\\theta=Tan^{-1}\dfrac{7.66m/s}{6.39m/s}\\\\\theta=50.17$

The lion leapt at an angle of 50.16°.

6 0

3 years ago