Please help on this one?

How do low energy electromagnetic waves compare with high energy electromagnetic waves? Select all

Dennis_Churaev [7]

Answer:

Explanation:

The different types of radiation are defined by the the amount of energy found in the photons. Radio waves have photons with low energies, microwave photons have a little more energy than radio waves, infrared photons have still more, then visible, ultraviolet, X-rays, and, the most energetic of all, gamma-rays

3 0

3 years ago

A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors

Mars2501 [29]

Answer:

(A) Capacitance per unit length = $4.02 \times 10^{-10}$

(B) The magnitude of charge on both conductor is $Q = 4.22 \times 10^{-19}$ C and the sign of charge on inner conductor is $+Q$ and the sign on outer conductor is $-Q$

Explanation:

Given :

Radius of inner part of conductor $(R_{1})$ = $2.7 \times 10^{-3}$ m

Radius of outer part of conductor $(R_{2})$ = $3.1 \times 10^{-3}$ m

The length of the capacitor $(l)$ = $3 \times 10^{-3}$ m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,

$C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }$

Where, $\epsilon_{o} = 8.85 \times 10^{-12}$

But we need capacitance per unit length so,

$\frac{C}{l} = \frac{2\pi\epsilon_{o} }{ln\frac{R_{2} }{R_{1} } }$

capacitance per unit length = $\frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}$

(B)

The charge on both conductors is given by,

$Q = C \Delta V$

Where, C = capacitance of cylindrical capacitor and value of $C = 12.06 \times 10^{-13}$ F, $\Delta V = 350 \times 10^{-3}$ V

∴ $Q = 4.22 \times 10^{-19}$ C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is $+Q$ and Charge on outer conductor is $-Q$ .

8 0

4 years ago

Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 10 cm in diameter if the me

lana66690 [7]

Answer:

dV = 32.98 cm³

Explanation:

We know that

V = f ( h,r)

$dV= \frac{\partial V}{\partial r}dr+ \frac{\partial V}{\partial h}dh$

We also know that

$V=\pi r^2h$

$dV= 2\pi h dr + \pi r^2 dh$

Given that

h= 10 cm

d=10 cm , r= 5 cm

Thickness = 0.05 cm so dr = 0.05 cm

The metal in the top and the bottom is 0.2 cm thick so dh = 0.2 + 0.2 cm

dh = 0.4 cm

Now by putting the values

$dV= 2\pi h dr + \pi r^2 dh$

$dV= 2\times \pi \times 5\times 0.05 + \pi \times 5^2\times 0.4\ cm^3$

dV = 32.98 cm³

8 0

3 years ago

Suppose there was a star with a parallax angle of 1 arcsecond. How far away would it be? Select all that apply.

lesya692 [45]

Answer:

option E

Explanation:

given,

Parallax angle(d) = 1 arcsecond

using Parallax formula

$d = \dfrac{1}{p}$

p is the parsecs angle which is measured in 1 arcsecond

d is the distance in parsec

now,

$P = \dfrac{1}{d}$

$P = \dfrac{1}{1}$

$P = 1 \ parsec$

we know,

1 parsec = 3.26 light year

hence, the answer will be option E

6 0

4 years ago

A Fourier analysis of the instantaneous value of a waveform can be represented by:

Viefleur [7K]

Answer:

t = 1

Explanation:

you dont know the value of the variable

7 0

3 years ago

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