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3 years ago

A light wave encounters a partial physical barrier, such as a wall with a hole in it. What is MOST LIKELY to occur?

2 answers:

8 0

The light will be diffracted.
diffraction - bending of a wave in a homogeneous medium around obstacles or through a narrow opening.

hope this helps :)

tatiyna3 years ago

5 0

Most likely, the light wave will be absorbed by the wall. Without any information as to the size and color of the wall, the location and size of the hole, or the location of the light wave, this is a generalized probability problem. For all of the places the light could be, it's more likely that it hits the wall than the hole (if the hole is less than 50% of the area of the wall).

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From the deepest to the surface, what are the parts of the earth's interior?

shepuryov [24]

The answer is A: Core --> Mantle --> Crust.

Core: The earth's core is the center of the earth, which would ultimately be the deepest. The core is made up of alloy, which is a mixture of many medals, such as iron and nickel.

Mantle: The earth's mantle is the layer between the earths crust and core. Often made of silicate rocks.

Crust: The earth's crust is the outer-most of the three options. Usually made of up different types of rocks.

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3 years ago

Help ASAP!! Three factors that affect the solubility of a substance are pressure, the type of solvent, and __________

romanna [79]

Answer:

Volume

Explanation:

7 0

3 years ago

Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

$f=k.x$

$\mu_s.N=k.x$

where:

N = m.g = the normal reaction force acting on the body under steady state.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

$U=\frac{1}{2} k.A^2$

This energy will be equal to the work done by the kinetic friction to stop it.

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

is the total distance does it travel before stopping.

7 0

3 years ago

A singer raises his pitch one octave. what quality of the sound wave has changed?

Keith_Richards [23]

I think the amplitude changed

3 0

3 years ago

The coefficients of friction between the load and the flatbed trailershown are μs = 0.40 and μk = 0.30. Knowing that the speed o

SOVA2 [1]

Answer:

50.97 m

Explanation:

m = Mass of truck

$\mu_s$ = Coefficient of static friction = 0.4

v = Final velocity = 0

u = Initial velocity = 72 km/h = $\dfrac{72}{3.6}=20\ \text{m/s}$

s = Displacement

Force applied

$F=ma$

Frictional force

$f=\mu_s mg$

Now these forces act opposite to each other so are equal. This is valid for the case when the load does not slide

$ma=\mu_s mg\\\Rightarrow a=\mu_s g\\\Rightarrow a=0.4\times 9.81\\\Rightarrow a=3.924\ \text{m/s}^2$

Since the obect will be decelerating the acceleration will be $-3.924\ \text{m/s}^2$

From the kinematic equations we have

$v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-20^2}{2\times -3.924}\\\Rightarrow s=50.97\ \text{m}$

So, the minimum distance at which the car will stop without making the load shift is 50.97 m.

5 0

3 years ago