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3 years ago

How is amplitude related to loudness

1 answer:

3 0

The loudness of a sound is linked to the size of the vibration which produces it. A big vibration makes a louder sound. Scientists use the word 'amplitude' for the size of waves. For waves on water, it is easy to measure the amplitude.

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Which of the following is a potential result of preparing appropriately before starting an experiment in a lab?

oksano4ka [1.4K]

Answer:

Your project goes well.

Explanation:

Because that's how it works.

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3 years ago

What real-world examples show no work being done? Can you think of examples other than resisting the force of gravity?

GalinKa [24]

-- pushing on a brick wall

-- standing on your little brother's back so that he can't get up

-- taking a nap while on the job

-- squeezing anything that doesn't yield to your squeeze, such as a glass bottle or your girl friend

-- watching TV

-- solving math problems in your head

-- making pictures out of clouds in the sky

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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3 years ago

Can i have the physical science grade 12 june 2021 paper (common paper)

Veseljchak [2.6K]

Answer:

u have to purchase it via online e-commerce platforms

7 0

3 years ago

Read 2 more answers

A 3.3 kg ball sits on the ground and is kicked with a FAPP of 36N

jeyben [28]

a) 32.3 N

The force of gravity (also called weight) on an object is given by

W = mg

where

m is the mass of the object

g is the acceleration of gravity

For the ball in the problem,

m = 3.3 kg

g = 9.8 m/s^2

Substituting, we find the force of gravity on the ball:

$W=(3.3)(9.8)=32.3 N$

b) 48.3 N

The force applied

$F_{app} = 36 N$

The ball is kicked with this force, so we can assume that the kick is horizontal.

This means that the applied force and the weight are perpendicular to each other. Therefore, we can find the net force by using Pythagorean's theorem:

$F=\sqrt{W^2+F_{app}^2}$

And substituting

W = 32.3 N

Fapp = 36 N

We find

$F=\sqrt{32.3^2+36^2}=48.3 N$

c) $14.6 m/s^2$

The ball's acceleration can be found by using Newton's second law, which states that

F = ma

where

F is the net force on an object

m is its mass

a is its acceleration

For the ball in this problem,

m = 3.3 kg

F = 48.3 N

Solving the equation for a, we find

$a=\frac{F}{m}=\frac{48.3}{3.3}=14.6 m/s^2$

8 0

3 years ago