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2 years ago

The speed of the pulse depends on what?

Physics

1 answer:

denis23 [38]2 years ago

8 0

Answer:

The pulse speed depends on the properties of the medium and not on the amplitude or pulse length of the pulse.

Explanation:

Hope this helps.

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Charra [1.4K]

Power = Work / time

The work given here is 83J and the time it took to do 83J of work was 3s

So..

Power = 83J / 3s

Power = 27.67 W or 27.7 W

8 0

3 years ago

When a rocket is traveling toward a mountain at 100 m/s, the sound waves from this rocket's engine approach the mountain at spee

Charra [1.4K]

Answer:

The correct option is C

Explanation:

From the question we are told that

The initial speed of the rocket is $v_i = 100 m/s$

The speed of the rocket engine sound is $V$

The final speed of the rocket is $v_f = 200 \ m/s$

The speed of the sound at $v_f$ would still remain V this because the speed of sound wave is constant and is not dependent on the speed of the observer(The mountain ) or the speed of the source (The rocket ).

A clear example when lightning strikes you will first see (that is because it travels at the speed of light which is greater than the speed of sound) but it would take some time before you hear the sound of the lightning

Here we see that the speed of the lightning(speed of sound) does not affect the speed of the sound it generates

3 0

2 years ago

2. A 20 cm object is placed 10cm in front of a convex lens of focal length 5cm. Calculate

adoni [48]

Answer:

 » Image distance :

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

v is image distance
u is object distance, u is 10 cm
f is focal length, f is 5 cm

${ \tt{ \frac{1}{v} + \frac{1}{10} = \frac{1}{5} }} \\ \\ { \tt{ \frac{1}{v} = \frac{1}{10} }} \\ \\ { \tt{v = 10}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: image \: distance \: is \: 10 \: cm \: \: }}}}}$

 » Magnification :

• Let's derive this formula from the lens formula:

${ \tt{ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} }} \\$

» Multiply throughout by fv

${ \tt{fv( \frac{1}{v} + \frac{1}{u} ) = fv( \frac{1}{f} )}} \\ \\ { \tt{ \frac{fv}{v} + \frac{fv}{u} = \frac{fv}{f} }} \\ \\ { \tt{f + f( \frac{v}{u} ) = v}}$

• But we know that, v/u is M

${ \tt{f + fM = v}} \\ { \tt{f(1 +M) = v }} \\ { \tt{1 +M = \frac{v}{f} }} \\ \\ { \boxed{ \mathfrak{formular : } \: { \tt{ M = \frac{v}{f} - 1 }}}}$

v is image distance, v is 10 cm
f is focal length, f is 5 cm
M is magnification.

${ \tt{M = \frac{10}{5} - 1 }} \\ \\ { \tt{M = 5 - 1}} \\ \\ { \underline{ \underline{ \pmb{ \red{ \: magnification \: is \: 4}}}}}$

 » Nature of Image :

Image is magnified
Image is erect or upright
Image is inverted
Image distance is identical to object distance.

4 0

1 year ago

Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons a

mrs_skeptik [129]

Answer:

Light of a shorter wavelength should be used.

Explanation:

This is studied in the phenomenon called photoelectric effect, in which light is able to release electrons from a metal, said electrons are called photoelectrons .

The experiments that have been carried out show that increasing or decreasing the intensity of the light will not cause the photoelectrons to be emitted, what will cause the photoelectrons to be emitted is to increase the frequency of the incident light.

And a higher frequency corresponds to a shorter wavelength according to the equation:

$f=\frac{c}{\lambda}$