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4 years ago

Calculate the wavelength of the electromagnetic radiation whose frequency is 7.5 x 10^12 Hz

Physics

1 answer:

tamaranim1 [39]4 years ago

4 0

velocity of electromagnetic waves=3.0×10^8.

wavelength=velocity/frequency

(3.0×10^8)/(7.5×10^12)

=0.00004

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Ilia_Sergeevich [38]

Use the formula,

$\Delta x=v_it+\dfrac12at^2$

where $\Delta x$ is the cart's displacement (from the origin), $v_i$ is its initial speed, $a$ is its acceleration, and $t$ is time.

$\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2$

$\implies\boxed{\Delta x=30\,\mathrm m}$

Alternatively, since acceleration is constant, we have

$\dfrac{v_f+v_i}2=\dfrac{\Delta x}t$

That is, we have these two equivalent expressions for average velocity, where $v_f$ is the cart's final velocity. Solve for $\Delta x$ :

$\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}$

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8 0

3 years ago

Is the image formed on the retina real or virtual

BabaBlast [244]

It is real, however it gets inverted in the retina and flips back in the cornea

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3 years ago

Read 2 more answers

A 0.290 cm diameter plastic sphere, used in a static electricity demonstration, has a uniformly distributed 30.0 pC charge on it

mojhsa [17]

Answer:

Therefore,

The potential (in V) near its surface is 186.13 Volt.

Explanation:

Given:

Diameter of sphere,

d= 0.29 cm

$radius=\dfrac{d}{2}=\dfrac{0.29}{2}=0.145\ cm$

$r = 0.145\ cm = 0.145\times 10^{-2}\ m$

Charge ,

$Q = 30.0\ pC=30\times 10^{-12}$

To Find:

Electric potential , V = ?

Solution:

Electric Potential at point surface is given as,

$V=\dfrac{1}{4\pi\epsilon_{0}}\times \dfrac{Q}{r}$

Where,

V= Electric potential,

ε0 = permeability free space = 8.85 × 10–12 F/m

Q = Charge

r = Radius

Substituting the values we get

$V=\dfrac{1}{4\times 3.14\times 8.85\times 10^{-12}}\times \dfrac{30\times 10^{-12}}{0.145\times 10^{-2}}$

$V=\dfrac{30}{16.117\times 10^{-2}}=186.13\ Volt$

Therefore,

The potential (in V) near its surface is 186.13 Volt.

3 0

4 years ago

Two loops of wire are arranged so that a changing current in one will induce a current in the other. If the current in the first

patriot [66]

Answer:

The current in the second loop will stay constant

Explanation:

Since the induced emf in the second coil, ε due to the changing current i₁ in the first wire loop ε = -Mdi₁/dt where M = mutual inductance of the coils and di₁/dt = rate of change of current in the first coil = + 1 A/s (positive since it is clockwise)

Now ε = i₂R where i₂ = current in second wire loop and R = resistance of second wire loop.

So, i₂R = -Mdi₁/dt

i₂ = -Mdi₁/dt/R

Since di₁/dt = + 1 A/s,

i₂ = -Mdi₁/dt/R

i₂ = -M × + 1 A/s/R

i₂ = -M/R

Since M and R are constant, this implies that i₂ = constant

<u>So, the current in the second wire loop will stay constant.</u>

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3 years ago

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