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3 years ago

Which occurs when two or more waves combine to produce a wave with a larger displacement?

Physics

2 answers:

Dimas [21]3 years ago

8 0

When two or more waves combine to produce a new wave, that's 'interference'.

-- If the new wave has larger displacements (amplitude), then it's <em>CON</em>structive interference.

-- If the new wave has smaller displacements (amplitude), then it's DEstructive interference.

slavikrds [6]3 years ago

7 0

Answer:

Due to constructive interference of two or more waves the net displacement is larger than the displacement of each wave.

Explanation:

When two or more waves superimpose at a point then the resultant displacement of medium particles is given by vector sum of displacement caused by each wave.

This phenomenon is known as superposition of waves and this is superposition of waves for coherent waves is known as interference.

Now in interference the net displacement is given by

$r = \sqrt{r_1^2 + r_2^2 + 2r_1r_2cos\theta}$

here

$r_1, r_2$ = displacement due to each wave

$\theta$ = phase difference between two waves

now if it gives constructive interference then they two waves must be in same phase and hence the resultant displacement must be more than the individual wave

$r = |r_1 + r_2|$

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leva [86]

Answer:

It would be hard to test scientifically since it's subjective and can only be proven true if you conducted some experimentations and observations.

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3 years ago

Six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then

Nataly_w [17]

Answer:

The option (b) is correct.

Explanation:

The expression for the power in terms of work done is as follows;

$P=\frac{W}{t}$

Here, W is the work done, t is the time taken and P is the power.

According to the given problem, six identical resistors are connected in series with a battery. The number of joules per second supplied by the battery is then determined.

The expression for the equivalent resistance in the series combination is as follows;

R_{eq}=R_{1}+R_{2}......+R_{6}

Put R_{1},R_{2}......,R_{6}=R.

R_{eq}=R+R+R+R+R+R

R_{eq}=6R

It is given in the problem that a seventh resistor is added (in series).

R_{eq}=R_{1}+R_{2}......+R_{7}

Put R_{1},R_{2}......,R_{7}=R.

R_{eq}=R+R+R+R+R+R+R

R_{eq}=7R

The expression for the power in terms of voltage and resistance is as follows;

$P=\frac{V^{2}}{R}$

Here, R is the resistance.

From the above expression, it can be concluded that the power, the number of joules per second supplied by the battery is inversely proportional to the resistance. The equivalent resistance increases if the seventh resistance is connected with a battery.

If a seventh resistor is added (in series) the number of joules per second supplied by the battery decreases.

Therefore, the option (b) is correct.

5 0

4 years ago

What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are

nexus9112 [7]

Answer:

Density of 18.0-karat gold mixture is $15.58 g/cm^3$ .

Explanation:

A mixture of 18 parts gold, 5 parts silver, and 1 part copper.

Let mass of gold be 18x

Let the mass of silver be 5x

Let the mass of copper be 1x

The density of gold = $19.32g/cm^3$

The density of silver = $10.1g/cm^3$

The density of copper = $8.8g/cm^3$

$Volume=\frac{Mass}{Density}$

Volume of the gold in the mixture = $V_1=\frac{18x}{19.32 g/cm^3}$

Volume of the silver in the mixture = $V_2=\frac{5x}{10.1 g/cm^3}$

Volume of the copper in the mixture = $V_3=\frac{1x}{8.8 g/cm^3}$

Mass of the mixture = M = 18x+5x+1x =24x

Volume of the mixture = $V_1+V_2+V_3$

Density of the mixture:

$\frac{M}{V_1+V_2+V_3}=15.58 g/cm^3$

8 0

3 years ago

The combustion of ethanol, C2H5OH(l), in oxygen to form carbon dioxide gas and water vapor is shown by which balanced chemical e

NemiM [27]

C2H5OH (l) + 3 O2 (g) = 2 CO2 (g)+ 3 H2O (l/g)

4 0

3 years ago

A series circuit has two 10-ohm bulb is added in a series. Technician A says that the three bulbs will be dimmer than when only

ANTONII [103]

Answer:

Technician A is right. The situation will happens even with only two bulbs in series

Explanation:

We must take into account that

1.- All electric device need its nominal voltage to operate

2.-Any and all electric device means an electric load for the source in terms of equation that means any device will implies a drop voltage of V = I*R ( I the flows current and R the resistance of the device)

3.-Nominal voltage for bulbs are specify for houses voltages you find between fase and neutral wires for instance in Venezuela 120 (v).

4.-In a imaginary circuit of only one bulb, the nominal voltage will be applied and the bulb will operates correctly, but when you add another bulb (in series) the nominal voltage will split between the two bulbs ( we could find a situation such as the first bulb work properly but the second one does not). The voltage split according to Ohms law (in such way that the sum of voltage between the terminal of the first bulb plus the voltage at terminals of the second one are equal to nominal voltage.

For that reason all the bulbs are connected in parallel in wich case all of them will operate with the common voltage

4 0

4 years ago