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3 years ago

If a wave hits a smooth surface at an angle of incidence of 40 degrees, the angle of reflection is

2 answers:

3 0

<span>D. 80 degrees.
</span>

The part of the ear where sound wave compressions and rarefactions cause the eardrum to vibrate is the middle ear. The 8th nerve in the inner ear actually converts the mechanical energy to electrical energy for transmitting to the brain. A membrane called the tympanic membrane separates the middle ear from the outer ear. Whenever a sound reaches the ear, it creates a sound wave that creates vibration in the eardrum. The pressure when high pushes the membrane inwards while low pressure sound waves helps the eardrum to come outwards.

icang [17]3 years ago

3 0

The angle of incidence is equal to the angle of reflection. If a wave hits a smooth surface at an angle of incidence of 40 degrees then the angle of reflection is 40 degrees.

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A stone is dropped from a tower 100 meters above the ground. The stone falls past ground level and into a well. It hits the wate

Elan Coil [88]

Make the base of the building zero. Then the initial distance is 100m, final distance unknown x. Use gravity, time and initial velocity to solve for final distance.
x - 100 = (0)(5) +(1/2)(-9.81)(5^2)
x - 100 = 0 - 122.625
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x = -22.625 m below ground

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3 years ago

In which of these samples do the molecules most likely have the most kinetic energy? (2 points)

Vinvika [58]

D, water vapor. Gaseous state would have more kinetic energy, they are moving faster. If you have to compare the same state, then higher temperature would have the higher kinetic energy. But if you have solid and liquid at the same temperature - then liquid would have more.

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3 years ago

Beers of snakes thunderstorms darkness in water are classified as blank phobias

Scrat [10]

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3 years ago

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

s344n2d4d5 [400]

Answer:

v = √2G $M_{earth}$ / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

Eo = K + U = ½ m1 v² - G m1 m2 / r1

Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

Eo = Ef

½ m1v² - G m1 $M_{earth}$ / R = - G m1 $M_{earth}$ / R

v² = 2G $M_{earth}$ (1 / R - 1 / Rinf)

If we do Rinf = infinity 1 / Rinf = 0

v = √2G $M_{earth}$ / R

Ef = = - G m1 m2 / R

The mechanical energy is conserved

Em = -G m1 $M_{earth}$ / R

Em = - G m1 $M_{earth}$ / R

R = int ⇒ Em = 0

6 0

3 years ago

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

3 years ago