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4 years ago

Which sound will most likely have the highest pitch?

Physics

2 answers:

olga55 [171]4 years ago

7 0

Answer: A whistle sound will most likely have the highest pitch

Explanation:

We know that,

Frequency is the pitch of the sound wave.

The frequency is high that means pitch is high and the frequency is low that means pitch is low of the wave.

Frequency = pitch = tone

We can explain the frequency as a pitch.

The singer uses the word for frequency is the pitch of the sound.

So, we can say that the highest frequency of a whistle is 1000 Hz that means the highest pitch is 1000 Hz.

Hence, A whistle sound will most likely have the highest pitch.

tankabanditka [31]4 years ago

6 0

its d A whistle (1000 Hz) on edge

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If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val

baherus [9]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ Using 1st equation of motion

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

$\\$

☯ Now, Finding the force exerted

$\\$

$\dashrightarrow\:\: \sf{F = ma}$

$\\$

$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ Hence, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

8 0

3 years ago

A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The

marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0

3 years ago

You have two lightweight metal spheres, each hanging from an insulating nylon thread. One of the spheres has a net negative char

kkurt [141]

Answer:attract each other

Explanation:

When two-sphere, one with a negative charge and another neutral is brought close together but do not touch then they try to attract each other.

This because of the polarization of the neutral sphere as it is placed in the vicinity of a negatively charged sphere. The negatively charged sphere will induce the positive charge in the neutral sphere and they will attract each other according to Columb law.

8 0

3 years ago

a parking lot is going to be 60m wide and 240m long which dimensions could be used please please help

Bezzdna [24]

Answer: The area of the parking lot is 14,400 meters squared.

Explanation:

We have the dimensions of the parking lot.

60m by 240m

The units used here are meters.

Now, if we want to know the area of the parking lot is equal to the product between the length and the width:

A = 60m*240m = 14,400 m^2

The area of the parking lot is 14,400 meters squared.

3 0

4 years ago

Read 2 more answers

During a certain comet’s orbit around the Sun, its closest distance to the Sun is 0.6 AU, and its farthest distance from the Sun

Ray Of Light [21]

Answer:

At the closest point

Explanation:

We can simply answer this question by applying Kepler's 2nd law of planetary motion.

It states that:

"A line connecting the center of the Sun to any other object orbiting around it (e.g. a comet) sweeps out equal areas in equal time intervals"

In this problem, we have a comet orbiting around the Sun:

- Its closest distance from the Sun is 0.6 AU

- Its farthest distance from the Sun is 35 AU

In order for Kepler's 2nd law to be valid, the line connecting the center of the Sun to the comet must move slower when the comet is farther away (because the area swept out is proportional to the product of the distance and of the velocity: $A\propto vr$ , therefore if r is larger, then v (velocity) must be lower).

On the other hand, when the the comet is closer to the Sun the line must move faster ( $A\propto vr$ , if r is smaller, v must be higher). Therefore, the comet's orbital velocity will be the largest at the closest distance to the Sun, 0.6 A.

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4 years ago