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natta225 [31]
3 years ago
12

I need help on this question please?​

Physics
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

3

Explanation:

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Two identical waves are moving in the same direction with the same speed. If the amplitude of the combination of the two waves i
umka2103 [35]

Answer:

The phase difference between these two waves is 141.1⁰

Explanation:

The displacement of the wave is given as;

Y = y_xSin(Kx - \omega t)+y_xSin(Kx- \omega t + \phi)\\\\Y = 2y_xCos(\frac{1}{2} \phi)Sine(Kx- \omega t + \frac{1}{2} \phi)

Amplitude, A = 2yₓCos(¹/₂Φ)

Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A

A = 2(1.5A)Cos(¹/₂Φ)

A = 3ACos(¹/₂Φ)

¹/₃ =  Cos(¹/₂Φ)

(¹/₂Φ) = Cos ⁻(0.3333)

(¹/₂Φ) = 70.55°

Φ = 141.1°

The phase difference between these two waves is 141.1⁰

4 0
3 years ago
Giving brainlist so plz help
Volgvan

Answer:

I DON KNOW

Explanation:

6 0
3 years ago
Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
fenix001 [56]

Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

b) E=3.04 \frac{GN}{C}

Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

which yields:

r_{AP}^{2}= 43 m^{2}

(I will assume the positions are in meters)

Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

so we substitute the values:

F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

which yields:

F_{AP}=418.14 MN

Now we can find its components:

F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

which yields:

F_{CP}=4.20 GN

Now we can find its components:

F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

F_{CPy}=-766.81 MN j

F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

7 0
4 years ago
What is the relation between acceleration due to gravity and radius of the earth? ​
emmasim [6.3K]

Answer:

As the earth is an oblate spheroid, its radius near the equator is more than its radius near poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

Formula: g = GM/r2

Dimensional Formula: M0L1T-2

Values of g in SI: 9.806 ms-2

Explanation:

Please Mark me brainliest

8 0
3 years ago
Read 2 more answers
La estrella mas proxima a la tierra esta a 2 años de luz. Calcula esta distancia en unidades del SI
tino4ka555 [31]

Answer:

2\ ly=1.89\times 10^{16}\ m

Explanation:

The question says that, "The closest star to the earth is 2 light years away. Calculate this distance in SI units".

Given that,

The distance between the closest star and the Earth is 2 light years.

The SI unit of distance is m. It means we need to convert light years to meters. We know that,

1\ ly=9.461\times 10^{15}\ m

2 light-years means,

2\ ly=2\times 9.461\times 10^{15}\ m\\\\=1.89\times 10^{16}\ m

So, the required distance is equal to 1.89\times 10^{16}\ m.

5 0
3 years ago
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