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3 years ago

Why do I hear a loud sound when I put my ear against the wall and knock

Physics

1 answer:

Salsk061 [2.6K]3 years ago

5 0

Explanation:

the reason you hear a loud sound when you knock on a wall and put your ear u against it is because when you knock on said wall vibrations are bouncing off of the wall there fore making that loud banging sound.

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A __ allows you to determine the position of an object <br><br>this is science plz helpp

Alex_Xolod [135]

The correct answer is C

8 0

3 years ago

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

Describe a situation in which you exert a force on something and it does not move. Identify the action force and the reaction fo

goldfiish [28.3K]

Answer:

Please find the answer in the explanation.

Explanation:

To describe a situation in which you exert a force on something and it does not move, you need to Identify the action force and the reaction force, the relative size of the forces, and the objects they act on.

The magnitude of the static friction between the object and the ground is larger than the magnitude of the force applied.

The object will not move if the magnitude of the static friction is greater than the magnitude of the force applied.

The action force is the force applied while the reaction force is equal to the frictional force between the object and the ground.

3 0

3 years ago

A penny is dropped from the Statue of Liberty

Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

$vf=vi+at$ ,

vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

$vf=0m/s+(9.8m/s^2)*0.30seconds$

Step 2: Solve

$vf=2.94m/s$

Use the equation below to solve for the displacement at 0.30 seconds

$x=vit+\frac{1}{2} at^{2}$

Step 1: Substitute the same variables with the knowns

$x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}$

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0

3 years ago

How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?

Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0

3 years ago