Ask question

Ask question

All categories

4 years ago

12

Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is

called a ( ) . The medium affects the speed of the wave that passes through it. One factor that affects the speed of a wave is the ( ) of medium. Some waves move faster in solids and some waves move faster in liquids and gases. Another factor that affects wave speed is the ( ) of the medium.

2 answers:

pantera1 [17]4 years ago

6 0

the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

vitfil [10]4 years ago

3 0

Answer:

medium, type, and temperature

Explanation:

You might be interested in

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

3 years ago

State some important uses of magnet

GarryVolchara [31]

Use of magnet:

Magnets are used in magnetic compass, doorbells, refrigerators.
Magnets are used in dynamos, motors, loudspeakers, microphones etc.
Ceramic magnets are used in computers.
Magnets are used in toys to give a magic effect.

3 0

3 years ago

Read 2 more answers

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

A sound wave is described by D(y,t)=(0.0200mm)× sin[(8.96rad/m)y+(3140rad/s)t+π/4rad], where y is in m and t is in s.

koban [17]

We have a wave function: D(y,t) and we want to know some things about it. 1. The direction the wave is travelling is negative y direction or -y. 2. Since sound waves are longitudinal waves, this sound wave is oscillating along the y axis. 3. The wavelength we can get from k=2π/λ, k is the wave number, λ is the wavelength. So λ=2π/k=6.28/8.96=0.7 m. 4. Before i get the wave speed i will calculate the period of oscillation. It can be calculated from: ω=2πf where ω is angular frequency and f is wave frequency. So f=ω/2π=3140/6.28=500 Hz and the period is T=1/f=1/500=0.002 s. 5. Wave speed is v=λ*f= 0.7*500=350 m/s.

4 0

3 years ago

Read 2 more answers

. While at the dentist’s office, Marion noticed the x-ray images of her teeth looked very

kkurt [141]

Answer:

B

Explanation:

7 0

3 years ago