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GrogVix [38]
3 years ago
8

Fill in the blank

Physics
2 answers:
elixir [45]3 years ago
8 0
Is there any types of answer to get an idea
Naya [18.7K]3 years ago
5 0
The reactivity of an alkali metal depends on the metals ability to lose electrons.
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Indicate whether the statement is true or false. A force can be simply defined as a push or a pull.
Sidana [21]
This B: False, because the definition is lacking.  

Force is when two objects interact with one another causing it to either move or not move. In our daily lives there are a lot of times force is exerted upon us, rather force is everywhere and here are the evidences:  

*Pushing a cart
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*Apple falling down from a tree.
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3 years ago
A bus travels east for 3 km, then north for 4 km. What is its final displacement?
ikadub [295]
5 km northeast. Left and up would make northeast

8 0
3 years ago
A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from
Lena [83]

Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s

Explanation:

I hope this helps☺️

8 0
3 years ago
Which are characteristics of electromagnetic waves?check all that apply.
emmainna [20.7K]

Correct choices are marked in bold:

travel in straight lines and can bounce off surfaces  --> TRUE, normally electromagnetic waves travel in straight lines, however they can be reflected by objects, bouncing off their surfaces

travel through space at the speed of light  --> TRUE, all electromagnetic waves in space (vacuum) travel at the speed of light, c=3\cdot 10^8 m/s)

travel only through matter  --> FALSE; electromagnetic waves can also travel through vacuum

travel only through space  --> FALSE, electromagnetic waves can also travel through matter

can bend around objects  --> TRUE, this is what happens for instance when diffraction occurs: electromagnetic waves are bended around obstacles or small slits

move by particles bumping into each other  --> FALSE, electromagnetic waves are oscillations of electric and magnetic fields, so no particles are involved

move by the interaction between an electric field and a magnetic field --> TRUE, electromagnetic waves consist of an electric field and a magnetic field oscillating in a direction perpendicular to the direction of motion of the wave

8 0
3 years ago
Read 2 more answers
A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut
blsea [12.9K]

Answer:

1.E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2.E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

\rho (r) = \alpha (1-\frac{r}{R})

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

\int\, da = 2\pi r h

where ‘h’ is the length of the imaginary Gaussian surface.

Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}

3. At the boundary where r = R:

E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}

As can be seen from above, two E-field values are equal as predicted.

4 0
3 years ago
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