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leva [86]
3 years ago
8

Describe the difference between a transverse wave and longitudinal wave. Include the parts that you would find on a transverse w

ave and longitudinal wave. Give an example of each.
PLEASE HELP! ITS A WRITTEN RESPONSE!
Physics
2 answers:
irina1246 [14]3 years ago
7 0

Answer:

A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.

Explanation:

hi

spayn [35]3 years ago
6 0
A transverse wave is a moving wave in which the current is perpendicular to the direction of the wave or path of propagation. A longitudinal wave are waves in which the displacement of the median is in the direction of the propagation.
Example:
Transverse- pond ripple
Longitudinal- crest and troff
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What is the relation between inertia and mass?
aleksandr82 [10.1K]

Answer:

The tendency of an object to resist changes in its state of motion varies with mass. Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. A more massive object has a greater tendency to resist changes in its state of motion.

Explanation:

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3 years ago
What is diffraction?
WITCHER [35]

Answer:

D) the bending of a wave as it moves around an obstacle or passes through a narrow opening

Explanation:

Even though light travels in straight line, it will bend around objects if their size is comparative to its wavelength. This phenomenon is called diffraction.

Light will also bend if the light travels from one medium to another medium at an angle, but that is called refraction.

Diffraction occurs in one medium. Refraction occurs when light travels from one medium to another.

5 0
3 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
Margaret [11]

Let  us consider two bodies having masses m and m' respectively.

Let they are  separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as -  F = G\frac{mm'}{r^{2} }   where G is the gravitational force constant.

From the above we see that F ∝ mm' and F\alpha \frac{1}{r^{2} }

Let the orbital radius of planet  A is r_{1}  = r and mass of planet is m_{1}.

Let the mass of central star is m .

Hence the gravitational force for planet A  is f_{1} =G \frac{m_{1}*m }{r^{2} }

For planet B the orbital radius  r_{2} =2r_{1} and mass m_{2} = 3 m_{1}

Hence the gravitational force f_{2} =G\frac{m m_{2} }{r^{2} }

                                                 f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }

                                                 = \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }

Hence the ratio is  \frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2}  }{Gmm_{1}/r_{1} ^2 }

                                      =\frac{3}{4}     [ ans]


                                                 

                           

3 0
3 years ago
Read 2 more answers
A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"
Mashutka [201]

The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g 


Then 0.5* 9.8m/s² * 18m = 88.2 would be v² 

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

8 0
3 years ago
Read 2 more answers
A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t
elena55 [62]

Answer:

The first part can be solved via conservation of energy.

mgh = mg2R + K\\K = mg(h-2R)

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}

where N = 0 because we are looking for the case where the car loses contact.

mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0
2 years ago
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