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2 years ago

Explain the difference between how a lamp and a tv use electrical energy

1 answer:

7 0

An authority style is a pioneer's style of giving guidance, executing arrangements, and rousing individuals. Different creators have proposed recognizing a wide range of initiative styles as displayed by pioneers in the political, business or different fields.

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The weight of the windsurfer is 700 newtons. Calculate the moment exerted by the windsurfer on the sailboat

borishaifa [10]

Hey, I think someone should help u cause I’m stuck too

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2 years ago

When a potassium atom forms an ion, it loses one electron. What is the electrical charge of the potassium ion? *

Ganezh [65]

+1 An electron has a negative charge so losing a charge of -1 from an uncharged, or neutral, atom will leave an ion with a positive charge.

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3 years ago

Read 2 more answers

What is the name of the chart that contain elements?

prisoha [69]

Answer:

Element Chart

Explanation:

It is a chart that provides many different elements.

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2 years ago

A student throws a rock upwards. The rock reaches a maximum height 2.4 seconds after it was released.

Allisa [31]

Answer:

23.52 m/s

Explanation:

The following data were obtained from the question:

Time taken (t) to reach the maximum height = 2.4 s

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =..?

At the maximum height, the final velocity (v) is zero. Thus, we can obtain how fast the rock (i.e initial velocity)

was thrown as follow:

v = u – gt (since the rock is going against gravity)

0 = u – (9.8 × 2.4)

0 = u – 23.52

Collect like terms

0 + 23.52 = u

u = 23.52 m/s

Therefore, the rock was thrown at a velocity of 23.52 m/s.

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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:

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

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

4 0

3 years ago