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exis [7]
2 years ago
6

A transverse wave is moving toward the right in a uniform medium. Point X represents a particle of the uniform medium. Which dia

gram represents the direction of the motion of particle X at the instant shown?

Physics
2 answers:
trapecia [35]2 years ago
6 0
After looking at the transverse waves in the diagram you listed above, the one diagram that does represent the direction of particle X at the instant show in diagram number 3. The direction of the wave motion is up. The correct answer choice will be 3. 
vladimir2022 [97]2 years ago
6 0
<h3><u>Answer</u>;</h3>

(4) last diagram

<h3><u>Explanation</u>;</h3>
  • <em><u>A transverse wave is a type of wave where the movement of the particles of the medium is perpendicular to the direction of the propagation of the wave. </u></em>
  • Transverse waves forms regions of maximum displacement known as crests and troughs. The crests are the top points of the wave and the trough is the bottom point of the wave.
  • In a transverse wave particles of the medium vibrate to and from in a direction perpendicular to the direction of energy transport.
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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s moti
vitfil [10]
Answer i think its d because the car is gradually going up
7 0
3 years ago
3. Magnetite is <br>a) Natural magnet <br>(b)Artificial magnet <br>(c) Not a magnet ​
andrey2020 [161]

\huge \bold \red  {Aɴswᴇʀ}

<em>Magnetite is a Natural magnet</em>

8 0
2 years ago
Read 2 more answers
Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

                     = 1.9858 x 10⁻²⁷ Kg

speed of light = 3 x 10⁸ m/s

binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

4 0
3 years ago
I need help with these questions :<br>(see image )​
dem82 [27]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2><h3>DATA:</h3>

Height of Aeroplane = 500 meters

Speed of Aeroplane = 200 m/s

Acceleration due to gravity = g = 10m/s^2

time of package to reach the ground = t = ?

Horizontal distance = d = ?

<h2>_____________________________________</h2><h3>SOLUTION:</h3>

When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the plane as the plane is just moving in horizontal direction. So the bomb fall and will travel forward.

Initial horizontal velocity is given by,

                                            V_{ix} = 200 ms{-1}

Initial vertical velocity is given by,

                                            V_{iy} = 0m/s

By the second equation of motion under gravity,

                                           H = V_{iy}t + \frac{1}{2}gt^2\\\\500 = (0)t + \frac{1}{2}x10xt^2 \\\\500 = 0 + 5t^2\\\\t^2 = \frac{500}{5}\\\\t^2 = 100\\\\t = 10 seconds

<h2>_____________________________________</h2>

<u>For horizontal distance(d)</u>:

Horizontal distance has no acceleration thus d is given by

                                          d = vt

                                          d = 200x10

                                          d = 2000 meters

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

                 

4 0
3 years ago
The diagram shows a ball resting at the top of a hill.
WARRIOR [948]

Answer:

a) The potential energy in the system is greatest at X.

Explanation:

Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}

Conclusions are showed as follows:

a) The potential energy in the system is greatest at X.

b) The kinetic energy is the lowest at X and Z.

c) Total energy remains constant as the ball moves from X to Y.

Hence, the correct answer is A.

7 0
3 years ago
Read 2 more answers
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