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svet-max [94.6K]

4 years ago

10

Look at the two wave diagrams. Which best describes the difference between wave A and wave B? Wave A has a greater pitch and fre

quency. Wave B has a greater pitch and frequency. Wave A has a greater pitch but lower frequency. Wave B has a lower pitch but greater frequency

2 answers:

djverab [1.8K]4 years ago

7 0

Answer:

A, Wave A has a greater pitch and frequency.

Explanation:

Did the test on edg!

stealth61 [152]4 years ago

4 0

Answer:

Correct answer is Wave A has a greater pitch and frequency

Explanation:

To understand the given figure, we should know the definition of wavelength, frequency and pitch

Wavelength

wavelength is defined as distance between two successive wave crests or troughs.

Frequency

Frequency is defined as the number of waves which passes through a given point per second. SI unit of frequency is Hertz.

Pitch

our ear understand the frequency of a sound wave as pitch. A a lower frequency sound has a lower pitch and a higher frequency sound has a higher pitch.

In given figure Wave A has lower wavelength and higher frequency so it have higher pitch than wave B.

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which of the following is not a property of an acid? reacts with metals bitter taste reacts with carbonate turns litmus paper re

notka56 [123]

The property that does not describe an acid is that it has a bitter taste. Acids have a sour taste, Bases have a bitter taste.

7 0

4 years ago

g a stone with mass m=1.60 kg IS thrown vertically upward into the air with an initial kinetic energy of 470 J. the drag force a

otez555 [7]

Answer:

Height reached will be 28.35 m

Explanation:

Here we can use the work energy theorem to find the maximum height

As we know by work energy theorem

Work done by gravity + work done by friction = change in kinetic energy

$-mgh - F_f h = 0 - \frac{1}{2}mv_i^2$

now we will have

$-1.60(9.8)(h) - 0.900(h) = - 470$

$-16.58 h = -470$

$h = 28.35 m$

so here the height raised by the stone will be 28.35 m from the ground after projection in upward direction

5 0

3 years ago

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above

Pachacha [2.7K]

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

$\theta=60^{\circ}$ is the angle of projection of the ball

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

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

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

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$

6 0

4 years ago

A student runs 35 m east, then 12 m west. What is the distance run by the student?

lora16 [44]

47m total just add them up 35 + 12 = 47

4 0

3 years ago

Read 2 more answers

In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

$V_{f}=a_{ave} \Delta t$

$V_{f}=(1.80 m/s^{2})(85.6 s)$

Finally:

$V_{f}=154.08 m/s$

8 0

3 years ago