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2 years ago

Nine steps for developing scientific principles are

Physics

1 answer:

xxTIMURxx [149]2 years ago

7 0

Answer:

1)Observe a phenomenon

2)Ask a question/ start inferring

3)Form a hypothesis

4)Create an experiment

5)Collect data

6)Compare results

7)Analyze

8)Report findings

9)Compare with other experiments

You might be interested in

Analyze how some insects are able to move around on the surface of a lake or pond

koban [17]

It is called surface tension it is the elastic personality of some liquids as they pull together to take up as little surface area as possible. the water molecules would rather stay together than be pulled apart<span />

3 0

3 years ago

A refrigerator removes 55.0 kcal of heat from the freezer and releases 73.5 kcal through the condenser on the back.How much work

sammy [17]

Here refrigerator removes 55 kcal heat from freezer

Refrigerator releases 73.5 kcal heat to surrounding

So here we can use energy conservation principle by II Law of thermodynamics

the law says that

$Q_1 = Q_2 + W$

here we know that

$Q_1$ = heat released to the surrounding

$Q_2$ = heat absorbed from freezer

W = work done by the compressor

now using above equation we can write

$73.5 = 55 + W$

$W = 73.5 - 55$

$W = 18.5 kcal$

So here compressor has to do 18.5 k cal work on it

5 0

3 years ago

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:

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