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2 years ago

Two wave pulses move toward each other along a rope.

Physics

2 answers:

mojhsa [17]2 years ago

7 0

Answer:

The answer is B. It is the one that has the triangle pointing up on the topside of the rope and seems to be a bit bigger than the other triangles.

Explanation:

lianna [129]2 years ago

3 0

Answer:

The second option with a vertex pointing upwards.

Explanation:

A wave is a disturbance that can be transmitted through a medium without causing a permanent displacement of the particles of the medium. The two types are; longitudinal and transverse.

In transverse wave, the direction of propagation of the wave is perpendicular to the direction of vibration of the particles of the medium. Examples are; water waves, light waves, X-ray etc. While in longitudinal waves, the direction of vibration of the particles is parallel to the direction of propagation of the waves. Examples are; waves in a slinky, waves in a rope, sound waves etc.

A wave in a rope is a form of longitudinal waves. When two wave pulses move toward each other, interference is observed.

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A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v

Pavel [41]

Answer:

Explanation:

The resonant frequencies for a fixed string is given by the formula nv/(2L).

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e fn+1 – fn = 390 -325

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3 years ago

if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given

yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given, $y=80t-16t^2$ gives the height in 't' seconds.

Average velocity = Rate of change of distance

= Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1) t = 0.01 s

y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by $y=80t-16t^2$ .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

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2 years ago