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3 years ago

State examples of a transverse wave.

Physics

2 answers:

blondinia [14]3 years ago

7 0

Answer

Light and other types of electromagnetic radiation are examples of transverse waves. Some other examples of transverse waves include a ripple on a pond and a wave in a string.

laiz [17]3 years ago

6 0

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

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Please help me on question 3a and 3b.<br><br>Thanks!

Sonbull [250]

(a) The frequency of water wave is 2 Hz.

(b) The wave speed of the water wave is 3.6 m/s.

<u>Explanation:</u>

(a) It is known that completion of one complete wave in 1 second is defined as frequency of 1 HZ. So here there are 120 waves crossing the boat in 1 minute. So the frequency of the water wave will be

$Frequency =\frac{\text { Number of waves }}{\text { Time in seconds }}$

As the time is 1 minute which is equal to 60 seconds and the number of waves is given as 120 then the frequency of the water wave is

$\text { Frequency }=\frac{120}{60}=2 \mathrm{Hz}$

So the frequency of water wave is 2 Hz.

(b) Then if the wavelength of the water wave is 1.8 m with a frequency of 2 Hz, then speed of the wave can be determined as the product of wavelength with frequency.

So Speed = Frequency × Wavelength

Speed = 2 × 1.8 = 3.6 m/s.

So the speed of the water wave is 3.6 m/s.

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3 years ago

A blender with a total resistance of 6 ohms runs on a voltage of 190 volts. How much current does it use? NEED TO KNOW ASAP PLEA

faltersainse [42]

<h2>Hello!</h2>

The answer is 31.67 Amps.

To solve the problem, we need to use Ohm's Law equation, which states that:

$V=IR$

Where,

V, is the voltage (in volts)

I, is the current (in Amps)

R, is the resistance (in Ohms)

We are given the following information:

$Voltage=190v\\Resistance=6\Omega\\$

So, using Ohm's Law equation and substituting the given information, we have:

$V=IR\\\\190v=I*6\Omega\\\\I=\frac{190v}{6\Omega}=31.67Amps$

Hence, we have that the current used by the blender is 31.67Amps.

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3 years ago

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slega [8]

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3 years ago

1. A car moves 20 m/s east for 5 sec, 10m/s north for 10 sec and then 30m/s west for 5 sec. Calculate a) displacement

aalyn [17]

Try the solutions described in the attached picture, note the answers are marked with green colour.

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3 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

3 years ago

State examples of a transverse wave. ​

State examples of a transverse wave.