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Please help giving 15 points<br> How do you calculate the density of an object?

Elis [28]

You first find the mass and the volume of that object. Then you divide mass ÷ volume

3 0

3 years ago

Read 2 more answers

Which statement correctly defines power?

Anna007 [38]

Answer:

$Power=V\, I$ which corresponds to the second option shown: "voltage times amperage"

Explanation:

The electric power is the work done to move a charge Q across a given difference of potential V per unit of time.

Since such electrical work is the product of the potential difference V times the charge that moves through that potential, and this work is to be calculated by the unit of time, we need to divide the product by time (t) which leads to the following final simple equation:

$Power=\frac{V\,Q}{t} =V\,\frac{Q}{t} = V\, I$

Notice that we replaced the quotient representing charge per unit of time (Q/t) by the actual current running through the circuit.

This corresponds to the second option shown in the question: "Voltage times amperage".

6 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:

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

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

4 0

3 years ago

List two methods by which a community can save water. In your own words, explain how these methods would help to increase the ye

Oliga [24]

Conservation and Rain Water Tank. Through conservation, they can save water in their own houses by lessening the use of water in an unimportant purpose; they can turn of the faucet when it is not used; they can use their laundry water to clean the bathroom and there are more. Through Rain Water Tank, they can create this structure in order to collect rain water and store it in the tank then use it afterwards such as in watering the plants.

8 0

3 years ago

An amateur astronomer looks at the moon through a telescope with a 15-cm-diameter objective. What is the minimum separation betw

Lana71 [14]

Answer:

y = 128.0 km

Explanation:

The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains

the diffraction equation for the first minimum is

a sin θ = λ

In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.

θ = 1.22 λ / d

d = 15 cm

to find the distance we can use trigonometry

tan θ = y / L

tan θ = sin θ / cos θ = θ

substituting

y / L = λ / d

y = L λ /d

let's calculate

y = 384 10⁸ 500 10⁻⁹ / 0.15

y = 1.28 10⁵ m

Let's reduce to km

y = 1.28 10⁵ m (1km / 10³ m)

y = 128.0 km

the correct answer is 120 km away

5 0

3 years ago