All categories

3 years ago

Sound travels at about 360 meters/sec. What is the wavelength of sound with a

Physics

1 answer:

ollegr [7]3 years ago

6 0

Answer:

0.24 m

Explanation:

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

We know:

v = λ / t [ f = 1 / t ]

v = λ f

= > λ = v / f

Putting values here we get:

= > λ = 360 / 1500 m

= > λ = 36 / 150 m

= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

You might be interested in

Write down short notes on: a. Electrical energy

PilotLPTM [1.2K]

Answer:

Explanation:

Electrical energy is energy derived from electric potential energy or kinetic energy.

Or,

Electrical energy is a form of energy resulting from the flow of electric charge. Lightning, batteries and even electric eels are examples of electrical energy.People use electricity for lighting, heating, cooling, and refrigeration and for operating appliances, computers, electronics, machinery, and public transportation systems.

Hope it helped you.

4 0

2 years ago

A beam of light, initially travelling in the air, strikes water surface at an angle of 24.5° with the normal. If the speed of li

kykrilka [37]

Answer:

Explanation:

n = 3.00e8/2.22e8 = 1.35

1.00sin24.5 = 1.35sinθ

θ = 17.9°

8 0

2 years ago

In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a

ElenaW [278]

In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a C. series of compressions and rarefactions.

7 0

2 years ago

Read 2 more answers

A wheel of moment of inertia of 5.00 kg∙m2 starts from rest and accelerates under a constant torque of 3.00 N∙m for 8.00 s. What

Strike441 [17]

Answer:

The wheel's rotational kinetic energy is 57.6 J.

Explanation:

Given that,

Moment of inertia = 5.00 kg.m²

Torque = 3.00 N.m

Time = 8.00 s

We need to calculate the angular acceleration

Using formula of the torque act on the wheel

$\tau=I\alpha$

$\alpha=\dfrac{\tau}{I}$

Where, I = moment of inertia

$\alpha$ = angular acceleration

$\tau$ = torque

Put the value into the formula

$\alpha=\dfrac{3.00}{5.00}$

$\alpha=0.6\ rad/s^2$

We need to calculate the final angular velocity

Initially wheel at rest so initial velocity is zero.

Using formula of angular velocity

$\alpha=\dfrac{\omega_{f}-\omega_{i}}{t}$

$\omega_{f}=\omega_{i}+\alpha t$

Put the value into the formula

$\omega_{f}=0+0.6\times8.00$

$\omega_{f}=4.8\ rad/s$

We need to calculate the rotational kinetic energy of the wheel

Using formula of the rotational kinetic energy

$K.E_{rot}=\dfrac{1}{2}I\omega^2$

$K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2$

$K.E_{rot}=57.6\ J$

Hence, The wheel's rotational kinetic energy is 57.6 J.

6 0

3 years ago

An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. The car travels to the bottom, through a loop, an

aalyn [17]

Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H) = (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) = 359,200 joules of energy.

There's no friction, so it'll have that same energy at every point of the story.
=================================

Skip the loop for a moment, because the first question concerns the hill after
the loop. We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = 40.71 meters
=================================

Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = 21.61 m/s
===============================

Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground, PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic. KE = (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = 28.63 m/s at the end
=======================================

If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere. Sorry.

6 0

2 years ago