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3 years ago

what will happen when two wave pulses that are the same size approach one another from opposite directions?

1 answer:

5 0

When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.

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The temperature of a substance is _________

emmainna [20.7K]

The answer is B.
Temperature is just another term for the average kinetic energy of a substance.

6 0

3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

3 years ago

What is the mechanical energy of a 500kg rollercoaster car moving with a speed of 3m/s at the top hill that is 30m high

koban [17]

K.E = 1/2*m*v^2 = 1/2(500)(3)^2 = 2250 J

m*g*h = 500(9.8)(30) = 147000 J

2250 + 147000 = 149250

4 0

3 years ago

Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210

kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light $(\lambda )=565nm \approx 565\times 10^{-9}m$

Width of slit(b)=0.210

(a)Width of central maximum located 1.80m from slit

$=\frac{2\lambda L}{b}$

$=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

=9.685 mm

(b)Width of the first order bright fringe

$Y_1=\frac{\lambda \times L}{b}$

$Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}$

$Y_1=4.84mm$

5 0

3 years ago

Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the s

Nadusha1986 [10]

Answer:

Explanation:

Let the velocity after first collision be v₁ and v₂ of car A and B . car A will bounce back .

velocity of approach = 1.5 - 0 = 1.5

velocity of separation = v₁ + v₂

coefficient of restitution = velocity of separation / velocity of approach

.8 = v₁ + v₂ / 1.5

v₁ + v₂ = 1.2

applying law of conservation of momentum

m x 1.5 + 0 = mv₂ - mv₁

1.5 = v₂ - v₁

adding two equation

2 v ₂= 2.7

v₂ = 1.35 m /s

v₁ = - .15 m / s

During second collision , B will collide with stationary A . Same process will apply in this case also. Let velocity of B and A after collision be v₃ and v₄.

For second collision ,

coefficient of restitution = velocity of separation / velocity of approach

.5 = v₃ + v₄ / 1.35

v₃ + v₄ = .675

applying law of conservation of momentum

m x 1.35 + 0 = mv₄ - mv₃

1.35 = v₄ - v₃

adding two equation

2 v ₄= 2.025

v₄ = 1.0125 m /s

v₃ = - 0 .3375 m / s

3 0

3 years ago