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3 years ago

Marcia pokes fun at her husband Brian because he loves to sing in the shower every morning, but

Physics

1 answer:

Masteriza [31]3 years ago

7 0

Answer:

Since I can see no choices, I answered it in my own understanding.

Brian - amplitude and frequency

Marcia - amplitude and longitudinal wave

Explanation:

"Sound" and "sound waves" are essential part of a person's life. They can be used for communicating and detecting some objects.

Brian loves singing in the shower which means that he is using a greater amplitude. Amplitude refers to the intensity of the sound or the amount of energy that a sound carries. When one sings in the shower, the sound cannot travel very far. It bounces immediately back to the person singing thus, making the sound bigger. Brian is also using a different range of frequency compared to his normal way of talking. The frequency of a normal male voice is normally 85 to 180 Hz. A person singing may have a frequency as high as 1,500 Hz.

Marcia talks loudly on the phone. This means that she is also using a greater amplitude because the intensity of her voice is big. Since she is using the telephone, this means that her voice travels in a longitudinal wave through the telephone. This allows her voice to reach to the person on the other end of the line.

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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)

T = 1.564s

So we use this in our formula:

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

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

2 years ago

A hockey stick of mass ms and length L is at rest on the ice (which is assumed to be frictionless). A puck with mass mp hits the

krek1111 [17]

Answer:

L = mp*v₀*(ms*D) / (ms + mp)

Explanation:

Given info

ms = mass of the hockey stick

uis = 0 (initial speed of the hockey stick before the collision)

xis = D (initial position of center of mass of the hockey stick before the collision)

mp = mass of the puck

uip = v₀ (initial speed of the puck before the collision)

xip = 0 (initial position of center of mass of the puck before the collision)

If we apply

Ycm = (ms*xis + mp*xip) / (ms + mp)

⇒ Ycm = (ms*D + mp*0) / (ms + mp)

⇒ Ycm = (ms*D) / (ms + mp)

Now, we can apply the equation

L = m*v*R

where m = mp

v = v₀

R = Ycm

then we have

L = mp*v₀*(ms*D) / (ms + mp)

5 0

2 years ago

The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el

Nana76 [90]

Answer:

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

and E = F/ q

substitute for F

E = ( K Qq/ d² ) / q

q cancel q

E = KQ / d²

so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²

7 0

3 years ago

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$