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3 years ago

8

Which of these illustrates that sound waves can travel through two different types of media? A. Talking to some one under the po

ol B. Hearing a loud whisper across the room C. Hearing people talking on the other side of the wall D. Hear thunder from a very distant storm

1 answer:

ankoles [38]3 years ago

4 0

Hearing people talking from the other side of the wall. This indicates sound travels in air and through the solid wall.

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No matter what values of m and k you used for the spring, what is the ratio of the period to k m.

Leto [7]

The relationship between the period of an oscillating spring and the attached mass determines the ratio of the period to $\sqrt{\dfrac{m}{k} }$ .

Response:

The ratio of the period to $\sqrt{\dfrac{m}{k} }$ is always approximately<u> 2·π : 1</u>

<u />

<h3>How is the value of the ratio of the period to $\sqrt{\dfrac{m}{k} }$ calculated?</h3>

Given:

The relationship between the period, <em>T</em>, the spring constant <em>k</em>, and the

mass attached to the spring <em>m</em> is presented as follows;

$T = \mathbf{2 \cdot \pi \cdot \sqrt{\dfrac{m}{k} }}$

Therefore, the fraction of of the period to $\sqrt{\dfrac{m}{k} }$ , is given as follows;

$\mathbf{\dfrac{T}{ \sqrt{\dfrac{m}{k} }}} = 2 \cdot \pi$

2·π ≈ 6.23

Therefore;

$T :{ \sqrt{\dfrac{m}{k} }} = 2 \cdot \pi : 1$

Which gives;

The ratio of the period to $\sqrt{\dfrac{m}{k} }$ is always approximately<u> 2·π : 1</u>

Learn more about the oscillations in spring here:

brainly.com/question/14510622

7 0

3 years ago

You observe two stars over the course of a year (or more) and find that both stars have measurable parallax. (1 arc second is 1/

Ierofanga [76]

Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

The distance between the Earth and the Sun is 150000000 Km. That distance is also known as an astronomical unit (1AU).

The parallax angle can be defined in the following way:

$\tan{p} = \frac{1AU}{d}$

Where d is the distance to the star.

$p('') = \frac{1}{d}$ (1)

Equation (1) can be rewritten in terms of d:

$d(pc) = \frac{1}{p('')}$ (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

$d(pc) = \frac{1}{1}$

$d(pc) = 1 pc$

So, star X is at 1 parsec from Earth.

Case of Star Y ( $p('') = \frac{1}{2}$ ):

$d(pc) = \frac{1}{(\frac{1}{2})}$

$d(pc) = 2 pc$

So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

5 0

3 years ago

I REALLY NEED HELP ON PHYSICS!!!<br> 10 POINTS

aleksley [76]

Answer:

25.021 sec

Explanation:

v=d/t

-2.33= - 58.3/t

t= - 58.3/-2.33

t=25.021 sec

6 0

3 years ago

Point A of the circular disk is at the angular position θ = 0 at time t = 0. The disk has angular velocity ω0 = 0.17 rad/s at t

statuscvo [17]

Given that,

Angular velocity = 0.17 rad/s

Angular acceleration = 1.3 rad/s²

Time = 1.7 s

We need to calculate the angular velocity

Using angular equation of motion

$\omega=\omega_{0}+\alpha t$

Put the value in the equation

$\omega=0.17+1.3\times1.7$

$\omega=2.38(k)\ m/s$

We need to calculate the angular displacement

Using angular equation of motion

$\theta=\theta_{0}+\omega_{0}t+\dfrac{\alpha t^2}{2}$

Put the value in the equation

$\theta=0+0.17\times1.7+\dfrac{1.3\times1.7^2}{2}$

$\theta=2.1675\times\dfrac{180}{\pi}$

$\theta= 124.18^{\circ}$

We need to calculate the velocity at point A

Using equation of motion

$v_{A}=v_{0}+\omega\times r$

Put the value into the formula

$v_{A}=0+2.38(k) \times0.2(\cos(124.18)i+\sin(124.18)j))$

$v_{A}=0.476\cos(124.18)j+0.476\sin(124.18)i$

$v_{A}=(-0.267j-0.393i)\ m/s$

We need to calculate the acceleration at point A

Using equation of motion

$a_{A}=a_{0}+\alpha\times r+\omega\times(\omega\times r)$

Put the value in the equation

$a_{A}=0+1.3(k)\times0.2(\cos(124.18)i+\sin(124.18)j)+2.38\times2.38\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=0.26\cos(124.18)i+0.26\sin(124.18)j+(2.38)^2\times0.2(\cos(124.18)i+\sin(124.18)j)$

$a_{A}=-0.146j-0.215i−0.636i+0.937j$

$a_{A}=0.791j-0.851i$

$a_{A}=-0.851i+0.791j\ m/s^2$

Hence, (a). The velocity at point A is $(-0.267j-0.393i)\ m/s$

(b). The acceleration at point A is $(-0.851i+0.791j)\ m/s^2$

3 0

4 years ago

A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori

Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>

</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.

That's the force that's accelerating the little block, so that must be the tension
in the string.

7 0

3 years ago