All categories

4 years ago

As a transverse wave travels through a medium, it displaces the particles perpendicular to the motion of the wave.

Physics

2 answers:

max2010maxim [7]4 years ago

5 0

Answer:

Perpendicular to the motion of wave.

Explanation:

Transverse waves are the waves that are defined as the waves whose oscillation is perpendicular to the motion of the wave.

An example of this motion is the string that produces waves by anchoring its one end and the other end will move up and down in this, the wave is propagated or its particles are moving perpendicular to the motion of the wave.

8_murik_8 [283]4 years ago

3 0

I am not 100% sure but my answer would be parallel to the direction of the wave motion.

You might be interested in

what will be the gravitational force between two bodies if the mass of each is doubled and the distance between them is halved?

katen-ka-za [31]

Answer:

Gravitational force will be 16 times more.

Explanation:

we know;

Gravitational force (F) = (Gm1m2)/d^2

when mass of each is doubled and distance between them is halved;

F= (G2m1×2m2)/(d/2)^2

=(4Gm1m2)/(d^2/4)

=4×4(Gm1m2)/d^2

=16(Gm1m2)/d^2

=16F

3 0

3 years ago

A block of mass 2 kg is launched by compressing a spring of force constant 1200 N/m. The block slides on a frictionless surface,

SVEN [57.7K]

Answer:

zzyibgsdwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwwe

Explanation:

3 0

3 years ago

The sound intensity from a jack hammer breaking concrete is 2.0W/m2 at a distance of 2.0 m from the point of impact. This is suf

kupik [55]

Answer:

(a) $I_{1}=3.2*10^{-3}W/m^{2}$

(b) $\beta =95dB$

Explanation:

Given data

Distance r₁=50 m

Distance r₂=2 m

Intensity I₂=2.0 W/m²

To find

(a) The Sound Intensity I₁

(b) The Sound Intensity level β

Solution

For (a) the Sound Intensity I₁

$\frac{I_{1} }{I_{2}}=\frac{(r_{2})^{2} }{(r_{1})^{2} }\\I_{1} =I_{2}(\frac{(r_{2})^{2} }{(r_{1})^{2} })\\I_{1}=(2.0W/m^{2} )(\frac{(2m)^{2} }{(50m)^{2} })\\I_{1}=3.2*10^{-3}W/m^{2}$

For (b) the Sound Intensity level β

The Sound Intensity level β is calculated as follow

$\beta =(10dB)log_{10}(\frac{I}{I_{o} } )\\\beta =(10dB)log_{10}(\frac{3.2*10^{-3}W/m^{2} }{1.0*10^{-12} W/m^{2} } )\\\beta =95dB$

8 0

3 years ago

A loudspeaker, mounted on a tall pole, is engineered to emit 75% of its sound energy into the forward hemisphere, 25% toward the

Naily [24]

Answer:

"0.049 W" is the correct answer.

Explanation:

According to the given question,

$r = \sqrt{(3.5)^2+(2.5)^2}$

$=\sqrt{8.5}$

$SL=85$

As we know,

⇒ $SL=10 \ log(\frac{I}{I_o} )$

$85=10 \ log(\frac{I}{10^{-12}} )$

$I=3.162\times 1^{-4} \ W/m^2$

Now,

⇒ $P_{front} = I(2\pi r^2)$

$=(3.162\times 10^{-4})(2\pi\times 18.5)$

$=0.0368 \ W$

$=0.75 \ P$

or,

$=0.049 \ W$

7 0

3 years ago

Which of the following statements indicates that the client is at risk for an electrical shock at home?

sashaice [31]

Answer:

“I had to cut off the third prong on the electrical plug so that it would fit in the extension cord.”

Explanation:

A good method to find the customer at greater risk of receiving an electric shock is to discard the options.

If we analyze the fourth option "I always make sure that I am standing in a dry area before operating electrical equipment.", we will see that the customer minimizes the risk of electric conduction through a fluid. This precaution also minimizes the risk of electric shock.

The third option "I always read the owner’s manual when I purchase a new electrical appliance." It tells us that the customer knows how to properly use the electrical equipment that manipulates and knows the do and do not of such devices, so the risk of electric shock is reduced.

In the second option, "My bread got stuck in my toaster this morning, and I unplugged it before trying to remove it." The customer made sure to interrupt the electric flow to the device before proceeding and try to remove the jammed bread. If there is no electric current, there will be no electric shock to the customer.

The first option "I had to cut off the third prong on the electrical plug so that it would fit in the extension cord." It tells us that the client altered the cord extension in an inappropriate manner, since the equipment is not designed to work with two prongs. Because the design conditions are not being met, it is possible that the equipment malfunctions and this malfunction could result in the customer receiving an electric shock.

I hope the explanation was clear for you. If you have any further question, I'll be happy to assist you. :D

6 0

4 years ago