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garri49 [273]
3 years ago
12

A light source radiates 60.0 W of single-wavelength sinusoidal light uniformly in all directions. What is the amplitude of the m

agnetic field of this light at a distance of 0.700 m from the bulb?
Physics
1 answer:
Anna35 [415]3 years ago
6 0

To solve this problem it is necessary to take into account the concepts of Intensity as a function of Power and the definition of magnetic field.

The intensity depending on the power is defined as

I = \frac{P}{4\pi r^2},

Where

P = Power

r = Radius

Replacing the values that we have,

I = \frac{60}{(4*\pi (0.7)^2)}

I = 9.75 Watt/m^2

The definition of intensity tells us that,

I = \frac{1}{2}\frac{B_o^2 c}{\mu}

Where,

B_0 =Magnetic field

\mu = Permeability constant

c = Speed velocity

Then replacing with our values we have,

9.75 = \frac{Bo^2 (3*10^8)}{(4\pi*10^{-7})}

Re-arrange to find the magnetic Field B_0

B_o = 2.86*10^{-7} T

Therefore the amplitude of the magnetic field of this light is B_o = 2.86*10^{-7} T

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0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for
Paladinen [302]

Assuming the accleration applied was constant, we have

v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)

\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)

\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}

Then the force applied to the ball is given by

F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)

\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N

8 0
3 years ago
A particle travels clockwise on a circular path of diameter​ R, monitored by a sensor on the circle at point​ P; the other endpo
kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}

With this function we should only calculate the derivate in function of c

\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}

That is the rate of change of \theta.

b) At this point we need only make a substitution of 0 for c in the equation previously found.

\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}

Hence we have finally the rate of change when c=0.

6 0
3 years ago
What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper
rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is 28.5cm^3

Explanation :

First we have to calculate the mass of copper.

\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}

\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g

Now we have to calculate the volume of copper.

Formula used :

Density=\frac{Mass}{Volume}

Now put all the given values in this formula, we get:

8.92\times 10^3kg/m^3=\frac{254g}{Volume}

Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3

Conversion used :

1kg/m^3=1g/L\\\\1L=10^3cm^3

Therefore, the volume of a sample of 4.00 mol of copper is 28.5cm^3

7 0
3 years ago
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Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley.Ifm1=2 kg and m2 = 3 kg, and bl
Vera_Pavlovna [14]

Answer:

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = (\frac{(m₂-m₁)}({m₁+m₂} * g = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

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Murrr4er [49]

Answer:

h=112.35

Explanation:

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