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Fofino [41]
2 years ago
14

Part A

Physics
1 answer:
inn [45]2 years ago
6 0

Answer:in 1905

Explanation:

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When water waves meet, they can combine to form new waves. In constructive waves, a ________ amplitude wave is formed. In destru
Sever21 [200]

Answer:

In constructive waves, a <u><em>greater</em></u> amplitude wave is formed. In destructive waves, a wave with a <u><em>smaller</em></u> amplitude is formed. (option A)

Explanation:

Interference is called the superposition or sum of two or more waves. Depending mainly on the wavelengths, amplitudes and the relative distance between them, there are two types of interference: constructive or destructive.

Constructive interference occurs when there are two waves of identical or similar frequency (both have motions equal to an even number of similar wavelengths) and overlap the peak of one with the peak of the other. These effects add together and make a wave of greater amplitude. All of this is possible because the waves were in the same phase in the beginning (in the same position).

Destructive interference occurs in the opposite case to constructive. When the crest of one wave overlaps the valley of the other, they cancel out since they are in different phases when they overlap (they were in different positions). That is, as in the case of constructive waves they were added, in the case of destructive waves they cancel out (subtract).

So, <u><em>In constructive waves, a greater amplitude wave is formed. In destructive waves, a wave with a smaller amplitude is formed. </em></u>

3 0
3 years ago
A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri
s344n2d4d5 [400]

Answer:

The\quad magnitude\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad \quad =2.52m/{ s }^{ 2 }\\ The\quad direction\quad of\quad the\quad car's\quad acceleration\quad at\quad t=13s\quad =15.{ 72 }^{\o}\\The\quad car\quad begins\quad to\quad slide\quad out\quad \quad of\quad the\quad circle\quad after\quad 26.09s.\quad \quad \quad \quad

Explanation:

8 0
3 years ago
A 1.50 3 103 - kg car starts from rest and accelerates uniformly to 18.0 m/s in 12.0 s. Assume that air resistance remains const
nexus9112 [7]

Answer:

Explanation:

let force exerted by engine be F.Net force =( F-400)N, applying newton law

     F-400 = 1.5 x 10³x18 =27000 ,

F = 27400 N.

velocity after 12 s  = 0 + 18 x 12 = 216 m/s

Average velocity = (0 + 216 )/2 = 108 m/s

Average power = force x average velocity = 27400 x 108 = 29.6 10⁵ W .⁶

b) At 12 s , velocity = 216 m/s

Instantaneous power = velocity x force = 216 x 27400 = 59.2 x 10⁶ W.

8 0
3 years ago
Read 2 more answers
a car accelerates uniformly from rest to a speed of 51.9mi/h in 9.37s. find the constant acceleration of the car. answer in unit
Darina [25.2K]
Here is my step-by-step-work. Let me know if you have any questions! :)

3 0
3 years ago
it takes 90 j of work to stretch a spring 0.2 m from its equilibrium position. How muc work is needed to stretch it an additiona
Vinvika [58]

Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

E=\frac{1}{2}kx^2

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J

Therefore, the additional work needed is

\Delta E=E'-E=810-90=720 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

7 0
3 years ago
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