The frequency of radio waves is inversely proportional to the wavelength. Given that the wavelength is 1.5 x 103 m when the frequency is 2.0 x 105 hertz (Hz), find The frequency of the radio waves with a wavelength of 480 m.( 10 marks )

1. The frequency of radio waves is inversely proportional to the wavelength. Given that the wavelength is 1.5 x 103 m when the frequency is 2.0 x 105 hertz (Hz), find The frequency of the radio waves with a wavelength of 480 m.( 10 marks )

Jawaban:

terjemah:

Frekuensi gelombang radio berbanding terbalik dengan panjang gelombang. Diketahui panjang gelombang 1,5 x 103 m pada frekuensi 2,0 x 105 hertz (Hz), tentukan frekuensi gelombang radio dengan panjang gelombang 480 m.( 10 tanda )

2. Find the inequalities system and find the maximum and minimum values for the objective quantity : z = x+ y​

Sub Chapter : Inequalities

Ways/Answer : Put on the Pict

3. how to calculate wavelength with distance and frequency

Answer

-

[tex]λ = \frac{v}{f} [/tex]

[tex]λ = v.T[/tex]

-

λ = wavelength

v = distance / fast wave

f = frequency

T = period

4. two forces of magnitude 3N and 7N respectively act on body. What are the maximum and minimum resultant forces that act on the body?​

Jawaban:

Max = 10 N

Min = 4 N

Penjelasan:

If there are two force act on the same body, the value will be maximum if both of them in same direction so we can add to calculate. Otherwise if one another isn't in same direction, it will be minimum (because we can subtract the big one with the little one)

5. The magnitudes of displacements and are 3 m and 4 m, respectively, and .Considering various orientations of and , what are (a) the maximum possible magnitude for and (b) the minimum possible magnitude?​

Jawaban:

Was it like this ? ; the question :

"The magnitudes of displacements a and b are 3m and 4m, and c=a+b. What is the maximum and minimum possible magnitude for c?"

Penjelasan:

if that's the case, then since displacement is a vector quantity and it follows vector addition so c=(a^2+b^2+2abcosA)^0.5

Where A is the angle between a and b

For maximum value of c A=0

c=a+b given a=3m b=4m

c=7m

For minimum value of c A=π

c=|a-b|

c=1m

Sorry....ifit's not, thentoobad. cz, Ican'tseem tounderstandyourquestion...

still, it's a goodexample. hopecanhelp you...

6. if x=30,35,40,45,50,55,60 and frequency=5,6,10,8,7,5,1 find the median

30 30 30 30 30 35 35 35 35 35 35 40 40 40 40 40 40 40 40 40 (40 45) 45 45 45 45 45 45 45 50 50 50 50 50 50 50 55 55 55 55 55 60

(40 + 45) ÷ 2

= 42,5

7. find the meaning of the following words! then arrange the frequency from the least​

Jawaban:

1. every time

2. in the way that most often happens

3. frequently

4. not often

5. rather than all of the time

6. at no time in the past or future

7. regularly or habitually

8. in most cases

9. not often

10. now and then

arrangements:

always,usually,generally,often,frequently, sometimes,occasionally, rarely,seldom,never

8. Find the periode and frequency of a simple pendulum 0,5 m long at a location where g = 9,8 m/s².

Penjelasan:

l = 0,5 m

g = 9,8 m/s²

T = ...?

f = ...?

jawab

[tex]T = 2\pi \sqrt{ \frac{l}{g} } \\ T = 2\pi \sqrt{ \frac{0.5}{9.8} } \\ T = 2\pi \sqrt{0.05} \\ T = 2\pi \times 0.224 \\ T = 0.448\pi[/tex]

T = 0,448π sekon

frekuensi

[tex]f = \frac{1}{T} \\ f = \frac{1}{0.448\pi} \\ f = \frac{2.232}{\pi} [/tex]

f = 2,232/π Hz

9. A guitar string vibrates with a frequency of 320 hz. If the wavelength of the vibration is 0.10 meters what is the speed of the wave?

Known:

f = 320 hz

λ = 0,1 m

Question:

v = ... m/s

Answer:

v = f x λ

v = 320 hz x 0,1 m

v = 32 m/s

The speed of the wave is 32 m/s

10. Find The Meaning of Following Advebs of Frequency!​

Jawaban:

1. Always: On every occasion

2. Usually: Most often

3. Often: Many times on different occasions

4. Rarely: Infrequently

5. Sometimes: Something happens on some occasions rather than all the time

6. Never: Not ever

7. Frequently: Something that occurs often or at frequent intervals

8. Every: The total number of people or things considered as a group

9. Seldom: On only a few occasions

10. Occasionaly: At infrequent or irregular intervals, now and then

Penjelasan:

11. The sum of two numbers x and y is 82. Find the maximum value of its p. Where p = xy

x + y = 82

nilai kali maksimal = 82/2 = 41

p = xy = (41×41) = 1681Bahasa Indonesia
untuk membuat hasil kali tertinggi
maka selisih x dan y harus sekecil mungkin
selisih terkecil = 0
berarti x = y
x + y = 82
x + x = 82
2x = 82
x = 82:2
   = 41
y = 41

P = xy
    = 41 x 41
    = 1681



Bahasa Inggris (In English)
we wanted to make the highest possible of the product
which means the difference of x and y must be as least as possible
smallest difference = 0
which means x = y
x + y = 82
x + x = 82
2x = 82
x = 82:2
   = 41
y = 41

P = xy
    = 41 x 41
    = 1681

12. from the weather forecast of big cities in the world,the maximum and the minimum temperature are:swiss:minimum -5 c and maximum 18 c ,bombay :minimum 17 c and maximum 34 c : paris:minimum -3 c and maximum 17 c ,Hongkong minimum -2 c and maximum 25 c. which city has the greatest temperature's differences

bombay and Paris city's

13. apa itu maximum dan minimum?

Penjelasan dengan langkah-langkah:

Maximum=Batas tertinggi

Minimum=Batas terendah

maaf kalo salah

Jawaban:

nilai minimum adalah fungsi objektif, yang menghasilkan nilai terendah pada daerah himpunan penyelesaian, misalnya biaya terendah. Nilai maksimum adalah fungsi objektif yang menghasilkan nilai tertinggi/ maksimum, misalnya laba, pendapatan.

semoga bermanfaat;)

14. A microwave oven uses microwaves with a wavelength of 0.12m. What is the frequency of these microwaves?

c = wavelength . f

3 . 10^8 = 0,12 . f

f = 2,5 . 10^9 Hz

15. Jelaskan bagaimana kesamaan WDM (Wavelength Division Multiplexing) dengan FDM (frequency Division Multiplexing)​

Jawaban:

Perbedaan antara FDM TDM dan WDM adalah bahwa FDM membagi bandwidth menjadi rentang frekuensi yang lebih kecil dan setiap pengguna mengirimkan data secara bersamaan melalui saluran umum dalam rentang frekuensi mereka, TDM mengalokasikan slot waktu tetap untuk setiap pengguna untuk mengirim sinyal melalui saluran umum ...

Penjelasan:

maap kalo kurang tepat

16. A siren Have speed 360 m/ s and frequency go Hz . How Much is it's wavelength

Dik : V = 360 m/s
          f  = 90 Hz

Dit : lamda = ___?

Penyelesaian : lamda = V / f
                                          = 360 m/s / 90 Hz
                                          = 4 m

== Semoga Membantu == #NoCopas
v=f λ
360=90 λ
λ=4 m

Semoga Membantu

17. Find the maximum and minimum point for y=3+4x³-x⁴. Sketch the graph of the curve.

Jawab:

Penjelasan dengan langkah-langkah:

[tex]y = 3+4x^3 - x^4\\\\y' = 12x^2 - 4x^3 = 0\\\\x^3 - 3x^2 = 0\\\\x^2(x-3) = 0\\\\x = 0 \;\cup\; 3\\\\y'' = 24x - 12x^2 = 0\\\\x^2 - 2x = 0\\\\x(x-2) = 0\\\\x = 0 \;\cup\; 2\\\\y''' = 24-24x = 0 => x = 1\\\\ y'''' = -24\\\\\begin{minipage}{15em}berdasarkan turunan keempat, ketiga, kedua, dan pertama :\\\\- f(x) terbuka keatas dan memiliki minimum pada interval $0 < x < 2$ \\\\- f(x) terbuka kebawah dan \\memiliki maksimum pada \\interval $x < 0$ atau $x > 2$ \\\\\end{minipage}[/tex]

[tex]\begin{minipage}{15 em}- f(x) memiliki perubahan kecekungan parabola singgung terbesar pada x = 1\\\\- berdasarkan turunan kedua maka f(x) memiliki maksimum lokal di x = 3\\\\- x = 0 pada interval $ 0< x < 2 $ bukan merupakan maksimum maupun minimum, tetapi secara keseluruhan tanpa interval x = 0 merupakan minimum f(x) (karena maksimum f(x) sudah ditemukan)\end{minipage}[/tex]

[tex]\begin{minipage}{15 em}- Nilai turunan keempat negatif dan mempengaruhi sifat turunan diatasnya \\- f(x) monoton turun ketika $x > 3$ dan monoton naik untuk \\$x < 0 \;\cap\; 0<x<3$ \\(netral pada x = 0)\end{minipage}[/tex]

maximum = f(3), "minimum" = f(0) = 3 = titik potong sumbu y

maximum = 3+4(3)³ - 3⁴  = 30

18. A traveling wave satisfy the equation y = 0,2 sin 0,4 π (60t – x). x and y in cm and t in seconds. Specify: a. amplitude wave b. wave frequency c. wavelength d. wave propagation

0,2 sin 0,4 phi (60t - x)
=> 0,2 sin 24 phi t - 0,4 phi x
=> A sin wt - kx

So :

A = 0,2
W = 24 phi t
k = 0,4 phi x

a.) Amplitude wave = 0,2 cm
b.) Wave frequency
w = 2 phi . f
24 phi = 2 phi . f
24/2 = f
12 Hz = f

c.) Wave length (lambda)

k = 2 phi / lambda
0,4 phi = 2 phi / lambda
Lambda = 2 / 0,4
Lambda = 5 cm

d.) Wave propagation (v)
v = Lambda . f
v = 5 . 12
v = 60 cm/s or 0,6 m/s

19. When the frequency of a wave is increased, what happens to :A.) the wavelength if the speed is constant?B.) The speed of the wave if the Wavelength is constant? Mohon bantuannya, hari ini duenya​

Penjelasan:

A. Wavelength = speed/frequency, so if frequency increased wavelength will decereased

B. Speed = wavelength. Frequency, so if frequency increased speed will increased

20. The ends of a tuning fork oscillate at a frequency of 440 Hz with an amplitude of 0.50 mm. Determine (a) the maximum velocity and (b) the maximum acceleration of the ends.​

(a) The maximum velocity of the ends can be determined using the equation:

v_max = A * ω

where A is the amplitude and ω is the angular frequency given by

ω = 2πf

Substituting the given values, we have:

v_max = 0.5 mm * (2π * 440 Hz) = 2828 mm/s

(b) The maximum acceleration of the ends can be determined using the equation:

a_max = ω^2 * A

Substituting the given values, we have:

a_max = (2π * 440 Hz)^2 * 0.5 mm = 3.12 x 10^7 mm/s^2

Note: The units for velocity and acceleration should be consistent, in this case mm/s and mm/s^2 respectively