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rodikova [14]
3 years ago
6

You are watching a wave with a frequency of 512 Hz go past. How many peaks of the wave go past you each minute?

Physics
2 answers:
bija089 [108]3 years ago
6 0

Answer:

61440

Explanation:

The period of a sine wave is the length of one cycle of the wave and is characterized by having two peaks a negative and a positive (See the image).

The frequency on the waves is given in hz equivalent to one cycle per second, so 512 hz are 512 cycles per second, given that one minute is equivalent to 60 seconds we have a total of 60*512 = 30720 cycles per second, because each cycle contains two peaks the total amount of peaks in one minute is 30720 *2 =61440

inna [77]3 years ago
4 0

Answer:

61440 peaks

Explanation:

A hertz represents one cycle for every second:

f= Hz=\frac{1}{s}

So if a wave have frequency of 2Hz for example, this means the wave does two cycles per second.

Normally the waves are represented by cosine or sine waves. These kind of waves have two peaks in each cycle, one positive and one negative. With this in mind, let's calculate how many peaks of the wave pass each minute.

A minute has 60 seconds, hence:

60*512=30720\hspace{3}cycles\hspace{3}per\hspace{3}minute

And we know already that every cycle has two peaks, so:

30720*2=61440\hspace{3}peaks\hspace{3}per\hspace{3}minute

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3 0
3 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

I =0.011 A

5 0
2 years ago
There is no air in space astronauts in space cannot hear sounds from outside their spacesuits explain this
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7 0
3 years ago
the half-life of iodine-131 is 8.1 days. how much time had passed if i only have one-fourth of the original sample?​
morpeh [17]

Answer:

16.2 days

Explanation:

Find the number of halflives:

1/2   *  1/2 =  1/4     so <u>two</u>   halflives have passed

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8 0
1 year ago
In the diagram, q1 = +6.60*10^-9 C and q2 = +3.10*10^-9 C. Find the magnitude of the total electric field at point P. pls help?
kirza4 [7]

Answer:

|E(t)| = 1258,46 [N/C]

α = 67,5⁰  (angle with respect x-axis)

Explanation:

E(t)  Electric Field is a vector, so we need to determine module and direction

E(t)  =  E(q₁)  + E (q₂)  Where E(q₁) and E (q₂) are electric fields due to electric charge q₁ and q₂  respectively.

E(q₁) = K * q₁/ (d₁)²         K = 9 *10⁹   [N*m²/C²]    d₁ = 0,350 m

E(q₁) = 9 *10⁹ * 6,6*10⁻⁹ / 0,1225      [N*m²/C²] *C/m²

E(q₁) = 484,9 [N/C]

E(q₂) =  9 *10⁹ * 3,1*10⁻⁹ / 0,024025

E(q₂) = 1161,29

Then

|E(t)| = √ |Eq₁|² + |Eq₂|²

|E(t)| = √ ( 484,9)² +( 1161,29)²

|E(t)| = √ 235128 + 1348594,46

|E(t)| = 1258,46 [N/C]

And tanα = 1161,29/484,9        tanα =  2,395      α = 67,5⁰

The angle of the vector electric field with the x-axis

3 0
3 years ago
Read 2 more answers
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