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Fantom [35]
2 years ago
9

Ten students stand in a circle and are told to make a transverse wave. What best describes the motion of the students? Each stud

ent bumps the shoulder of his or her neighbor. At the exact time, all students take a step to the right. The students skip clockwise, going in a circular motion. One at a time, each student lifts his or her hands up and then down.
Physics
2 answers:
ELEN [110]2 years ago
6 0

Answer:

The correct option here is D.

<em>~Hope this helps! :)</em>

Bond [772]2 years ago
3 0

Lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students. Option D is correct.

<h3>What is a Transverse wave?</h3>
  • The wave in which the oscillation of particles is is perpendicular to the direction of energy transfer.

  • The students can make a transverse wave by raising their hands up and then down, one student at a time.

  • The raised hand represents the oscillation of particles while the sequence of the raising hand represents the direction of energy transfer.

Therefore, lifting hands and the down by one student at a time best describe the presentation of the transverse wave by students.

Learn more about Transverse waves:  

brainly.com/question/3813804

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A 1000 kg box is being pushed with a force of 3500 N. What acceleration is the
WARRIOR [948]
Net Force = mass x acceleration
3500=1,000a
So a= 3500/1000
a=35/10
a=3.5 m/s^2
4 0
3 years ago
We only see objects because they absorb light. <br> True or False?
Alenkasestr [34]

Answer:

true

Explanation:

i think it's true because I took a quiz on this

3 0
3 years ago
2. A car is moving with a constant speed of 10m/s. What is the total distance the car travels
Tems11 [23]

Answer:

1200 meters

Explanation:

there are 60 seconds in a minute times 2 is 120 ten times 120 is 1200

6 0
3 years ago
From a window that is 20 m from the ground a stone with a speed of 10m / s is thrown vertically upwards. Calculate:
Oduvanchick [21]

a)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the maximum height

v = final velocity at the maximum height = 0 m/s

t = time taken to reach the maximum height

Using the equation

v² = v₀² + 2 a (Y - Y₀)

0² = 10² + 2 (- 9.8) (Y - 20)

Y = 25.1 m


also using the equation

v = v₀ + a t

inserting the values

0 = 10 + (- 9.8) t

t = 1.02 sec


b)

consider the motion in upward direction as positive and down direction as negative

Y₀ = initial position of the stone = 20 m

v₀ = initial velocity of the stone = 10 m/s

a = acceleration = - 9.8 m/s²

Y = final position of the stone when it reach the ground = 0 m

t = time taken to reach the ground

Using the equation

Y = Y₀ + v₀ t + (0.5) a t²

0 = 20 + 10 t + (0.5) (- 9.8) t²

t = 3.3 sec

3 0
3 years ago
A mass moves back and forth in simple harmonic motion with amplitude A and period T.
Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

1 T : 4 A = t : 2 A

and solving for t we find

t=\frac{(1T)(2 A)}{4A}=0.5 T

b. 1.25T

Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that

- the mass takes a time of 1 T to cover a distance of 4A

we can set the following proportion:

1 T : 4 A = t : 5 A

And by solving for t, we find

t=\frac{(1T)(5 A)}{4A}=\frac{5}{4} T=1.25 T

6 0
3 years ago
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