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Licemer1 [7]
2 years ago
9

Charina says that when waves interact with an object, they will interfere with the object, and when waves interact with other wa

ves, they will reflect off each other. do you agree with her? why or why not?
Physics
2 answers:
qaws [65]2 years ago
7 0

I completely don't agree with her, as a result of interference happens once waves act with alternative waves. Reflection happens once waves act with an associated object and bounce off it.

Explanation:

No, she has it backward. Waves interfere with one another and mirror off objects. once 2 waves overlap their amplitudes add. If they need a similar sign this addition is constructive, which means the amplitudes grow. If they need opposite signs this constitutes subtraction and also the waves will partly, or utterly cancel. this can be referred to as interference.

Reflection happens once waves travel from one medium to a different. If the wave electric resistance of the new medium is totally different (which it usually is) there'll be a partial, or maybe total, reflection.

Fofino [41]2 years ago
3 0
I would not agree with her since reflection and refraction happens only when waves hit an object. When, waves meet it is either it experiences constructive or destructive interference. Hope this answers the question. Have a nice day.
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The equation for gear ratio
Ymorist [56]
In a gear train with two gears, the gear ratio is defined as follows
R= \frac{\omega _A}{\omega _B} 

where \omega _A is the angular velocity of the input gear while \omega _B is the angular velocity of the output gear. 

This can be rewritten as a function of the number of teeth of the gears. In fact, the angular velocity of a gear is inversely proportional to the radius r of the gear:
\omega = \frac{v}{r}
But the radius is proportional to the number of teeth N of the gear. Therefore we can rewrite the gear ratio also as
R= \frac{\omega _A}{\omega _B} = \frac{r_B}{r_A} = \frac{N_B}{N_A}

4 0
3 years ago
"a musical tone sounded on a piano has a frequency of 261.6 hz and a wavelength of 1.31 m. what is the speed of the sound wave
Andreyy89
To solve this question, we use the wave equation which is:
C=f*λ
where:
C is the speed;
f is the frequency;
λ is the wavelength
So in this case, plugging in our values in the problem. This will give us:
C = 261.6Hz × 1.31m
= 342.696 m/s is the answer.
7 0
3 years ago
Read 2 more answers
An airplane is moving at 350 km/hr. If a bomb is
bogdanovich [222]

Answers:

a) -171.402 m/s  

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final height

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb's initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's final velocity

Knowing this, let's begin with the answers:

<h3>b) Time </h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity </h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign only indicates the direction is downwards

<h3>c) Range </h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
A tennis ball is served horizontally from 2.4m above the ground at net is 12m away and point 0.9 high will be ball clear the net
Schach [20]

Explanation:

Let us first calculate  long does it take to go 12m at 30m/s( assumed speed)

12/30 = 0.4 seconds

horizontal distance the ball drop in that time

H= (0)(0.4)+1/2(-9.8)(0.4)2

H= -0.78m

negative sign shows that the height of the ball at the net from the top.

Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m

As 1.62m>0.9m so the ball will clear the net.

H_1= V0y t’ + ½ g t’^2

-2.4= (0)t’ + ½ (-9.8) t’^2

t’= 0.69s

X’=V0x t’

X’=(30)(0.96)

X’= 20.7m

3 0
2 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
2 years ago
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