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Nimfa-mama [501]

3 years ago

6

Monochromatic electromagnetic radiation illuminates an area of a surface. The electric and magnetic fields of the waves are then

doubled in magnitude. How is the total energy incident on the surface per unit time affected by this increase in the electric and magnetic fields?
a) The total energy is not affected by this change.
b) The total energy will increase by a factor of two.
c) The total energy will increase by a factor of four.
d) The total energy will decrease by a factor of two.
e) The total energy will decrease by a factor of four.

1 answer:

emmasim [6.3K]3 years ago

6 0

Answer:

c) True. factor four

Explanation:

The energy density of an electromagnetic wave is given by

u = ½ ε₀ E² = B² / 2μ₀

Where ε₀ is the dielectric constant and μ₀ the magnetic permittivity.

Let's apply this equation to the present case

If we double the electro field

E ’= 2 E₀

u’= ½ ε₀ (2E₀)²

u’= ½ ε₀ Eo² 4

u’= 4 u₀

Therefore the energy is multiplied by four

Let's check the answers

a) False

b) False

c) True

d) False

e) False

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Answer:

200 km\h

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When a mass M hangs from a vertical wire of length L, waves travel on this wire with a speed V. What will be the speed of these

Zigmanuir [339]

Answer:

a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀

Explanation:

The speed of a wave along an eta string given by the expression

v = $\sqrt{ \frac{T}{ \mu } }$

where T is the tension of the string and μ is linear density

a) the mass of the cable is double

m = 2m₀

let's find the new linear density

μ = m / l

iinitial density

μ₀ = m₀ / l

final density

μ = 2m₀ / lo

μ = 2 μ₀

we substitute in the equation for the velocity

initial v₀ = $\sqrt{ \frac{T_o}{ \mu_o} }$

with the new dough

v = $\sqrt{ \frac{T_o}{ 2 \mu_o} }$

v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }

v = 1 /√2 v₀

v = 0.7071 v₀

b) we double the length of the cable

If the cable also increases its mass, the relationship is maintained

μ = μ₀

in this case the speed does not change

c) the cable l = l₀ and m = 3m₀

we look for the density

μ = 3m₀ / l₀

μ = 3 m₀/l₀

μ = 3 μ₀

v = $\sqrt{ \frac{T_o}{ 3 \mu_o} }$

v = 1 /√3 v₀

v = 0.577 v₀

d) l = 2l₀

μ = m₀ / 2l₀

μ = μ₀/ 2

v = $\sqrt{ \frac{T_o}{ \frac{ \mu_o}{2} } }$

v = √2 v₀

v = 1.41 v₀

e) m = 10m₀ and l = 2l₀

we look for the density

μ = 10 m₀/2l₀

μ = 5 μ₀

we look for speed

v = $\sqrt{ \frac{T_o}{5 \mu_o} }$

v = 1 /√5 v₀

v = 0.447 v₀

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Why do some scientists think that Irr II galaxies have irregular, distorted shapes?

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3 years ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

A satellite of mass m is in a circular orbit of radius R2 around a spherical planet of radius R1 made of a material with density

Amiraneli [1.4K]

Answer:

a) K = 2/3 π G m ρ R₁³ / R₂ , b) U = - G m M / r

Explanation:

The law of universal gravitation is

F = G m M / r²

Part A

Let's use Newton's second law

F = m a

The acceleration is centripetal

a = v² / R₂

G m M / R₂² = m v² / R₂

v² = G M / R₂

They give us the density of the planet

ρ = M / V

V = 4/3 π R₁³

M = ρ V

M = ρ 4/3 π R₁³

v² = 4/3 π G ρ R₁³ / R₂

K = ½ m v²

K = ½ m (4/3 π G ρ R₁³ / R₂)

K = 2/3 π G m ρ R₁³ / R₂

Part B

Potential energy and strength are related

F = - dU / dr

∫ dU = - ∫ F. dr

The force was directed towards the center and the vector r outwards therefore there is an angle of 180º between the two cos 180 = -1

U- U₀ = G m M ∫ dr / r²

U - U₀ = G m M (- r⁻¹)

We evaluate for

U - U₀ = -G m M (1 / $r_{f}$ - 1 / $r_{i}$ )

They indicate that for ri = ∞ U₀ = 0

U = - G m M / r

6 0

3 years ago