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4 years ago

Click any that apply. Examples of heat conduction via radiation include:

Physics

2 answers:

Ahat [919]4 years ago

6 0

Answer:

The sun and a heat lamp.

2 and 4

trapecia [35]4 years ago

3 0

The sun and heat lamp I think I could be wrong

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What is cos-^1(0.34)?<br> A. 19.9<br> B. 44.2°<br> C. 70.1°<br> D. 18.8°

Fudgin [204]

A calculator must be used. To put your calculator in degree mode, press the MODE button and select degree, the press the 2nd button then MODE again. For most TI calculators, press the 2nd button then press the cos button then enter the value 0.34. This will give you an answer of 70.123 (when you round to 3 decimal places).

The answer is C

3 0

3 years ago

A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c

Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

d = m1λD/β1

and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

y = 2β2 = 2 ( m2 / m1 ) β1

put here value

y = 2 ( 2 / 1 ) 0.7

y = 2.8 cm

so distance between the two second-order minima is 2.8 cm

6 0

3 years ago

Reviewing old tests and quizzes will not help you determine what will be on a test because teachers don’t like to give students

kifflom [539]

Answer:

False

Explanation:

5 0

4 years ago

Read 2 more answers

From Gauss's law, the electric field set up by a uniform line of charge is given by the following expression where is a unit vec

Evgesh-ka [11]

Answer:

$\Delta V=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

Explanation:

Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:

$2\pi r L*E=\lambda *L/\epsilon_{o}$

$E=\lambda /\(2 \pi* r *\epsilon_{o})$

We calculate the potential difference from the electric field:

$\Delta V=-\int\limits^{r_{1}}_{r_{2}} E \, dr =-\int\limits^{r_{1}}_{r_{2}} \lambda dr/ (2\pi*r*\epsilon_{o})=\lambda *ln(r_{2}/r_{1}) /\ (2\pi*\epsilon_{o})$

5 0

4 years ago

Electronic signals are converted into a ____ when transmitted through fiber-optic cables.

Nikolay [14]

Electronic signals are converted into a 'ray of light' -

3 0

4 years ago