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kobusy [5.1K]
3 years ago
13

If one wave of 10m amplitude meets constructively with another wave of amplitude 20m, what is the new amplitude?

Physics
1 answer:
Soloha48 [4]3 years ago
8 0
The answer is 30m.  <span>If one wave of 10m amplitude meets constructively with another wave of amplitude 20m, the new amplitude is 30m.  </span>When wave<span>  interfere </span>constructively<span>, their </span>amplitudes<span> are added. So the </span>amplitude<span> in this case is </span>10 m+20 m<span>=30 m .</span>
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natita [175]

Answer:

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Explanation:

3 0
3 years ago
Give an example of a vertical motion with a positive velocity and a negative acceleration. Give an example of a vertical motion
Assoli18 [71]

Answer:

An example of positive velocity is throwing a ball upwards

An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height

Explanation:

In a vertical movement the acceleration is always downwards, therefore negative since it is created by the attraction of the Earth on the body.

An example of positive velocity is throwing a ball upwards

An example of downward vertical velocity is when an object is dropped, for example a ball dropped from a height

7 0
4 years ago
A 20.0 Ω, 15.0 Ω, and 7.00 Ω resistor are connected in parallel to an emf source. A current of 7.00 A is in the 15.0 Ω resistor.
Lady bird [3.3K]

PART A)

Equivalent resistance in parallel is given as

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}

now we have

\frac{1}{R} = \frac{1}{20} + \frac{1}{15} + \frac{1}{7}

R = 3.85 ohm

PART B)

since potential difference across all resistance will remain same as all are in parallel

so here we can use ohm's law

V = iR

As we know i = 7 A current flows through 15 ohm resistance

V = (7 A)(15 ohm) = 105 volts

PART C)

Similarly ohm's law for 20 ohm resistance we can say

V = iR

105 = i(20 ohm)

i = 5.25 A

3 0
4 years ago
Read 2 more answers
When hydrogen is heated or subjected to an electric discharge, it emits light with specific visible wavelengths. A diffraction g
BartSMP [9]

To solve this problem we will apply the concepts related to the double slit-experiment. For which we will relate the distance between the Slits and the Diffraction Angle with the order of the bright fringe and the wavelength, this is mathematically given as,

d sin\theta = m\lambda

Here,

d = Distance between Slits

m = Order of the fringes

\lambda = Wavelength

\text{Spacing between adjacent lines} = d = 4926nm

\text{Wavelength of light} = \lambda = 656nm

Rearranging to find the angle,

sin\theta = \frac{m\lambda}{d}

\theta = sin^{-1}(\frac{m\lambda}{d})

\theta = sin^{-1}(\frac{(4)(656*10^{-9}m)}{4926*10^{-9}m})

\theta = 32.19\°

Therefore the angle that the fourth order bright fringe occur for this specific wavelenth of light occur is 32.19°

3 0
4 years ago
Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

1\times 500+\frac{(200)^2}{25037}+0=1\times T_2+\frac{(1700)^2}{25037}

500+1.598=T_2+115.4

T_2=500+1.598-115.4

T_2=386.2^{\circ}F

7 0
4 years ago
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