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Ray Of Light [21]

2 years ago

9

Last week Tasha and Amy got into a fight at school. Tasha is accusing Amy of talking behind her back. Since the fight, Tasha has

been
spamming Amy's social media profile with posts calling her a backstabber, a liar, and a cheat, among other things, Amy insists the matter was
all a misunderstanding, but some of her friends have started to turn against her.
In this situation Amy should try to confront Tasha face to face to sort the matter out. (5 points)
O True
O False

2 answers:

Leno4ka [110]2 years ago

5 0

Answer:

True

Explanation:

stepladder [879]2 years ago

4 0

Answer:

True

Explanation:

I feel that if she tries sorting it out herself that it could go two ways.

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Piano tuners tune pianos by listening to the beats between the harmonics of two different strings. When properly tuned, the note

Sophie [7]

(a) 2 Hz

The frequency of the nth-harmonic is given by

$f_n = n f_1$

where

$f_1$ is the fundamental frequency

Therefore, the frequency of the third harmonic of the A ( $f_1 = 440 Hz$ ) is

$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

$f_2 = 2 \cdot f_1 = 2 \cdot 659 Hz =1318 Hz$

So the frequency difference is

$\Delta f = 1320 Hz - 1318 Hz = 2 Hz$

(b) 2 Hz

The beat frequency between two harmonics of different frequencies f, f' is given by

$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

$f_1 \propto \sqrt{T}$

this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,

$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

8 0

2 years ago

A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system,

NARA [144]

As per the question the initial speed of the car [ u] is 42 m/s.

The car applied its brake and comes to rest after 5.5 second.

The final velocity [v] of the car will be zero.

From the equation of kinematics we know that

$v=u+at$ [ here a stands for acceleration]

$0=42 +5.5a$

$a =\frac{-42}{5.5} m/s^2$

$a= -7.64 m/s^2$

Here a is taken negative as it the car is decelerating uniformly.

We are asked to calculate the stopping distance .

From equation of kinematics we know that

$S=ut+\frac{1}{2} at^2$ [here S is the distance]

$= 42*5.5 +\frac{1}{2} [-7.64] [5.5]^2 m$

$=115.445 m$ [ans]

8 0

2 years ago

Select a musical instrument. Research the instrument and explain how it works in terms of sound production and resonance. Descri

iren [92.7K]

1. I think a drum would be the easiest instrument to explain. You can vary the pitch and loudness by hitting it with different intensities or at different points or angles.

PS. I'm not talking about a drumset. I'm talking about an ordinary drum.

2. Different types of instruments sound different even though they play the same notes because they use different mechanisms for producing sound.

7 0

3 years ago

Read 2 more answers

On topographic maps, contour lines that are farther apart indicate what ?

gladu [14]

Answer:

if I am correct, they indicate less steep terrain. think of it as the steeper the terrain the closer together the lines would be. hope that makes sense for you guys.

4 0

2 years ago

In 1935, a French destroyer, La Terrible, attained one of the fastest speeds for any standard warship. Suppose it took 3.0 min a

REY [17]

Answer:

Explanation:

From newton's equation of motion of uniform acceleration

v = u + at

where v is final velocity , u is initial velocity , a is acceleration and time is t .

putting the values

v = 0 + .5 x 3 x 60 ( time in second = 3 x 60 s )

= 90 m /s

So , final velocity is 90 m /s .

7 0

3 years ago