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4 years ago

Please help!!................

Physics

1 answer:

Gnoma [55]4 years ago

3 0

D. When a substance reacts with another substance, it shows its chemical property

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A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil

krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

Emf induced in the coil represented by formula

∈ = $-N\frac{d\phi}{dt}$ ...................(1)

Where:

. $\phi = BAcos\theta$ { B is magnetic field }

{A is cross-sectional area}

. $N =$ No. of turns in coil.

. $\frac{d\phi}{dt} =$ Rate change of induced Emf.

Here,

Considering the case :-

$N1 = 2N$ & $\frac{d\phi1}{dt} = 2\frac{d\phi}{dt}$

Putting these value in the equation (1) and finding the new emf induced (∈1)

∈1 = $-N1\times\frac{d\phi1}{dt}$

∈1 = $-2N\times2\frac{d\phi}{dt}$

∈1 = $4 [-N\times\frac{d\phi}{dt}]$

∈1 = 4∈ ...............{from Equation (1)}

Hence,

The Resultant Induced Emf in coil is 4∈.

8 0

4 years ago

An easy 20 points + brainiest.

Papessa [141]

Answer:

Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, and Neptune, Milky Way ;-;

4 0

2 years ago

In a cyclotron, the orbital radius of protons with energy 300 keV is 16.0 cm . You are redesigning the cyclotron to be used inst

dem82 [27]

Answer:

$r_\alpha=16cm$

Explanation:

The radius of the circumference described by a particle in a cyclotron is given by:

$r=\frac{mv}{qB}(1)$

m is the particle's mass, v is the speed of the particle, q is the particle's charge and B is the magnitude of the magnetic field.

Kinetic energy is defined as:

$K=\frac{mv^2}{2}=\frac{m^2v^2}{2m}\\$

Solving this for mv:

$mv=\sqrt{2mK}(2)$

Replacing (2) in (1):

$r=\frac{\sqrt{2mK}}{qB}$

For protons, we have:

$r_p=\frac{\sqrt{2m_pK}}{eB}(3)$

For alpha particles, we have:

$r_\alpha=\frac{\sqrt{2m_\alpha K}}{(2e)B}(4)$

Dividing (4) in (3):

$\frac{r_\alpha}{r_p}=\frac{\frac{\sqrt{2m_\alpha K}}{(2e)B}}{\frac{\sqrt{2m_p K}}{(e)B}}\\r_\alpha=\frac{r_p}{2}\sqrt{\frac{m_\alpha}{m_p}}\\r_\alpha=\frac{16cm}{2}\sqrt{\frac{6.64*10^{-27}kg}{1.67*10^{-27}kg}}\\r_\alpha=\frac{16cm}{2}(\sqrt{3.98})\\\\r_\alpha=\frac{16cm}{2}(2)\\r_\alpha=16cm$

4 0

4 years ago

What is the frequency corresponding to a period of 8.01 s?

masha68 [24]

The frequency of an oscillation is equal to the reciprocal of the period:
$f= \frac{1}{T}$
where
f is the frequency
T is the period

In our problem, the period is T=8.01 s, therefore the corresponding frequency is
$f= \frac{1}{8.01 s}=0.12 s^{-1} = 0.12 Hz$

8 0

3 years ago

You need to produce a solenoid that has an inductance of 2.47 μh . you construct the solenoid by uniformly winding 1.15 m of thi

Neporo4naja [7]

Answer:

0.0170m or 1.70cm

Explanation:

L=(μ₀*N^2*A)/(l)

N=x/2*pi*r so r=x/2*pi*N

A=pi*r^2=(pi*x^2)/(4*pi^2*N^2)

L=(μ₀*N^2*A)/(l)=(μ₀*N^2*pi*x^2)/(4*pi^2*N^2*l)=(μ₀*x^2)/(4*pi*l)

l=(μ₀*x^2)/4*pi*L)=[(4*10^-7)*(1.15m)/4*pi*2.47*10^-6H)=0.0170m or 1.70cm

Hope this helps. Any questions please feel free to ask. Thanks!

3 0

3 years ago