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3 years ago

14

An electromagnetic wave in vacuum has an electric field amplitude of 470 V/m. Calculate the amplitude of the corresponding magne

tic field.

1 answer:

lapo4ka [179]3 years ago

6 0

Answer:

Magnetic field, B = $B=1.56\times 10^{-6}\ T$

Explanation:

It is given that,

The amplitude of an electromagnetic wave, E = 470 V/m

We need to find the amplitude of the corresponding magnetic field. The relation between electric and magnetic field is :

$B=\dfrac{E}{c}$

Where

c is the speed of light

$B=\dfrac{470\ V/m}{3\times 10^8\ m/s}$

B = 0.00000156

$B=1.56\times 10^{-6}\ T$

So, the amplitude of the corresponding magnetic field is $1.56\times 10^{-6}\ T$ . Hence, this is the required solution.

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Please answer this question for me and explain why.

horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass $m_2$ and the 4kg mass $m_1$ . If the tension in the string is $T$ then for the mass $m_2$

(1). $T-m_2g=-m_2a$ <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass $m_1$

(2). $T-m_1g =m_1a$ <em> (the acceleration is upwards, hence the positive sign)</em>

Solving for $T$ in the 2nd equation we get:

$T =m_1a+m_1g$ ,

and putting this into the 1st equation we get:

$m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g$

$\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}$

Back to the question:

Using the formula for the acceleration we find

$a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g$

$a = \dfrac{g}{9},$

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

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3 years ago

A wire carries a steady current of 2.20

alina1380 [7]

By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.

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3 years ago

Glycerin (C3H8O3) is a nonvolatile liquid. What is the vapor pressure of a solution made by adding 154 g of glycerin to 316 mL o

garri49 [273]

Answer:

$P_{sol}=50.4\ mm.Hg$

Explanation:

According to given:

molecular mass of glycerin, $M_g=3\times 12+8+3\times 16=92\ g.mol^{-1}$

molecular mass of water, $M_w=2+16=18\ g.mol^{-1}$

∵Density of water is $0.992\ g.cm^{-3}= 0.992\ g.mL^{-1}$

∴mass of water in 316 mL, $m_w=316\times 0.992=313.5 g$

mass of glycerin, $m_g=154\ g$

pressure of mixture, $P_x=55.32\ torr= 55.32\ mm.Hg$

temperature of mixture, $T_x=40^{\circ}C$

<em>Upon the formation of solution the vapour pressure will be reduced since we have one component of solution as non-volatile.</em>

<u>moles of water in the given quantity:</u>

$n_w=\frac{m_w}{M_w}$

$n_w=\frac{313.5}{18}$

$n_w=17.42 moles$

<u>moles of glycerin in the given quantity:</u>

$n_g=\frac{m_g}{M_g}$

$n_g=\frac{154}{92}$

$n_g=1.674 moles$

<u>Now the mole fraction of water:</u>

$X_w=\frac{n_w}{n_w+n_g}$

$X_w=\frac{17.42}{17.42+1.674}$

$X_w=0.912$

<em>Since glycerin is non-volatile in nature so the vapor pressure of the resulting solution will be due to water only.</em>

$\therefore P_{sol}=X_w\times P_x$

$\therefore P_{sol}=0.912\times 55.32$

$P_{sol}=50.4\ mm.Hg$

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It is weight, if I understand your question.

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