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kirza4 [7]
3 years ago
5

What is occurring when two waves traveling along the same medium meet and cancel each other out?

Physics
1 answer:
liberstina [14]3 years ago
5 0

Answer:

Destructive interference

Explanation:

Destructive interference occurs when two waves traveling along the same medium meet and cancel each other out. The resultant wave has a smaller amplitude than the individual waves. In this case, the maxima of two waves are 180 degrees out of phase.

Hence, the correct option is (A) "destructive interference"

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The distance between the crests of waves
stepladder [879]
It depends, You have to have the length and the width of the crest wave.
8 0
3 years ago
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A 0.50 kg object is attached to a spring with force constant 157 N/m so that the object is allowed to move on a horizontal frict
Genrish500 [490]

Explanation:

We have,

Mass of an object is 0.5 kg

Force constant of the spring is 157 N/m

The object is released from rest when the spring is compressed 0.19 m.

(A) The force acting on the object is given by :

F = kx

F=157\times 0.19\\\\F=29.83\ N

(B) The force is simply given by :

F = ma

a is acceleration at that instant

a=\dfrac{F}{m}\\\\a=\dfrac{29.83}{0.5}\\\\a=59.66\ m/s^2

6 0
3 years ago
A circular dartboard has a radius of 2 meters and a red circle in the center. Assume you hit the target at a random point. For w
Sati [7]

Answer:

1.549 m

Explanation:

Given:

The radius of the circular board, r = 2 m

The probability of hitting the red is given as 0.6

Now, this probability of hitting the red can be conclude as

0.6 = (Area of red)/ (Total area of the board)

Total area of the board = πr² = π × 2²

let the radius of the red area be R

thus, area of red circle, = πR²

on substituting the value of the area, we have

0.6 = (πR²)/ (π × 2²)

or

R² = 2.4

or

R = 1.549 m

Thus, the radius of the red circle is 1.549 m

3 0
3 years ago
How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners
Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

U =\frac{Kq_{1}q_{2}}{d}

So, the total potential energy of teh system is

U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

U =3\times \frac{Kq^{2}}{d}

K = 9 x 10^9 Nm^2/C^2

By substituting the values

U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}

U = 1.38 x 10^-18 J

6 0
4 years ago
A particle oscillates in simple harmonic motion, with amplitude A and period T. The particle starts from position x = A. What is
frozen [14]

Answer:

i guess it is D because,

y = A sin(wt)

w = \frac{2\pi }{T} and t = 3T/ 2

now, Y = A sin ( \frac{2\pi }{T} X \frac{3T}{2})

so, Y = 0radian

Explanation:

8 0
3 years ago
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