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alexdok [17]
3 years ago
14

Tara prepared a report to show how the amplitude of waves affects the energy of waves. Is her graphical representation correct?

Physics
2 answers:
irga5000 [103]3 years ago
8 0

No, it is not correct because as the amplitude of a wave increases, the energy it carries also increases. This graph shows that as the amplitude increases, the energy decreases.                  

Pls mark as brain-list

Gelneren [198K]3 years ago
6 0

Answer:

It is not correct because the amplitude of the waves can be bigger than others and the graph can be going up and down

Explanation: I got the question right

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Regions in the milky way where density waves have caused gas clouds to crash into each other are called
Mandarinka [93]
Regions in the milky way where density waves have caused gas clouds to crash into each other are called clumps.Clumps are molecular clouds (interstellar clouds) with higher density,where lots of dust and gs cores resides. These clouds are the beginning of stars.
7 0
3 years ago
A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a
SOVA2 [1]

Answer:

a)    x = v₀² sin 2θ / g

b)    t_total = 2 v₀ sin θ / g

c)    x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

         v₀ₓ = v₀ cos θ

         v_{oy} = 13.5 sin 32 = 7.15 m / s

         v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

         v₀ₓ = x / t

          x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

          v_{y} = v_{oy} - gt

          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

           t_total = 2 t

           t_total = 2 v₀ sin θ / g

c) we calculate

          x = 13.5 2 sin (2 32) / 9.8

          x = 16.7 m

5 0
3 years ago
A toy car accelerates from 3m/s to 5m/s in 5sec . what is the acceleration​
san4es73 [151]

Answer:

0.4

Explanation:

Because 3m/s is the initial velocity(u) and 5m/s is the final velocity(v) and time is 5 sec.

So, acceleration = v-u ÷ t

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7 0
3 years ago
A rocket, initially at rest on the ground, accelerates straight upward from rest with constant acceleration 58.8m/s2 . The accel
Zanzabum
Split the operation in two parts. Part A) constant acceleration 58.8m/s^2, Part B) free fall.

Part A)
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Now you need the final speed to use it as initial speed of the next part.

Vf = Vo + at = 0 + 58.8m/s^2 * 7.00 s = 411.6 m/s

Part B) Free fall

Maximum height, y max ==> Vf = 0

Vf = Vo - gt ==> t = [Vo - Vf]/g = 411.6 m/s / 9.8 m/s^2 = 42 s

ymax = yo + Vo*t - g[t^2] / 2

ymax = 1440.6 m + 411.6m/s * 42 s - 9.8m/s^2 * [42s]^2 /2
ymax = 1440.6 m + 17287.2m - 8643.6m = 10084.2 m

Answer: ymax = 10084.2m
8 0
3 years ago
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