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Doss [256]
3 years ago
9

At

Physics
1 answer:
AnnyKZ [126]3 years ago
4 0

Explanation:

One common property with all halogens in group 7 is that they are all non-metals.

Fluorine, chlorine, Bromine and Iodine are classified as non-metallic elements and they have a high electronegativity.

  • In chemical reactions, they are very reactive because they require just one electron to complete their octet configuration and be isoelectronic with noble gases.

The most prominent observation from halogens is that they all non-metallic in nature.

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if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be
Kisachek [45]

Answer:

I_{rms}=2.12\ A

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

I=3\sin\theta\ A

We need to find the rms value of the current.

The general equation of current is given by :

I=I_o\sin\theta

It means, I_o=3\ A

We know that,

I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A

So, the rms value of current is 2.12 A.

4 0
3 years ago
People cannot see certain types of light waves because:
enot [183]

Answer:

Light comes in different colors like radio, ultra violet, gamma-ray, etc, and they are invisible to the bare eye

Explanation:

5 0
2 years ago
A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 3.20 cm i + 4.30 cm j. The segme
myrzilka [38]

The component of the force in negative z-direction is -0.144 N.

The given parameters;

  • <em>current in the wire, I = 2.7 A</em>
  • <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
  • <em>magnetic filed, B = 1.24 i</em>

The force on the segment of the wire is calculated as follows;

F = ILBsin(\theta)

where;

  • <em>θ is the angle wire and magnetic field</em>

<em />

The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.

The angle between the wire and the magnetic field is calculated as follows;

\theta = tan^{-1} (\frac{y}{x} )\\\\\theta = tan^{-1} (\frac{4.3}{3.2} )\\\\\theta = 53.3 \ ^0

The magnitude of the wire length is calculated as follows;

|l | = \sqrt{3.2^2 + 4.3^2} = 5.36 \ cm = 0.0536 \ m

The component of the force in negative z-direction is calculated as;

F_z = -ILB sin(\theta)\\\\F_z = -2.7 \times 0.0536 \times 1.24 \times  sin(53.3)\\\\F_z = -0.144 \ N

Thus, the component of the force in negative z-direction is -0.144 N.

Learn more here:brainly.com/question/22719779

6 0
3 years ago
A motorist travels at 270 km in p hours at a certain average speed. an expression for the distance that this motorist travelled
wel
Average speed = total distance / time ⇒ total distance = average speed *  time

Average speed = 270 km / p hours
distance = d
hours = x

d = 270/p * x
5 0
3 years ago
A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing throu
Setler79 [48]
<h2>Given :</h2>

  • total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

1.6 \times 10 {}^{ - 19} \:  C

For producing 1 Cuolomb charge we need :

  • \mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }

  • \dfrac{10 {}^{19} }{1.6}

  • \dfrac{10\times 10 {}^{19} }{16}

  • \dfrac{100 \times 10 {}^{18} }{16}

  • \mathrm{6.24 \times 10 {}^{18}  \:  \: electrons}

Now, for producing 0.009 C of charge, the number of electrons required is :

  • 0.009 \times 6.24 \times  {10}^{18}

  • 0.05616 \times 10 {}^{18}

  • \mathrm{5.616 \times 10 {}^{16}  \:  \: electons}

_____________________________

So, Number of electrons passing through the cross section in 3.6 seconds is :

\mathrm{5.616 \times 10 {}^{16} \:  \: electrons}

Number of electrons passing through it in 1 Second is :

  • \dfrac{5.616 \times  {10}^{16} }{3.6}

  • \mathrm{1.56 \times 10 {}^{16}  \:  \: electrons}

Now, in 10 seconds the number of electrons passing through it is :

  • 10 \times  \mathrm{1.56 \times 10 {}^{16}  \:  \: }

  • \mathrm{1.56 \times 10 {}^{17}  \:  \: electrons}

_____________________________

\mathrm{ \#TeeNForeveR}

6 0
3 years ago
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