What heavy element was the chernobyl nuclear accelarators supposed to make?

An 11.0 -W energy-efficient fluorescent lightbulb is designed to produce the same illumination as a conventional 40.0-W incandes

yarga [219]

The amount or cost that the user of the energy-efficient bulb save during 100h of use will be $0.319.

<h3>How to calculate the cost?</h3>

For the 11.0W bulb, it should be noted that the value will be:

= 11.0 × 100 × (1/1000) × 0.110

= $0.121

The 40W bulb will be:

= 40 × 100 × (1/1000) × 0.110

= $0.44

Therefore, the amount that will be saved will be:

= $0.44 - $0.121

= $0.319

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6 0

2 years ago

5. You head downstream on a river in an outboard.

Elena-2011 [213]

Answer:

hope this helps you're welcome

5 0

3 years ago

Where should you place the values of the independent variable when constructing a data table?

kobusy [5.1K]

<span>On the y-axis (the bottom of the table) hope this helps</span>

4 0

3 years ago

Read 2 more answers

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

3 years ago

Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Consider also a straight line throu

Sedbober [7]

Answer:

$\vec{V} = \frac{\Gamma}{2R}\vec{A}$

Explanation:

We define our values according to the text,

R= Radius

$\vec{V} =$ Velocity

$\Gamma =$ Strenght of the vortex filament

From this and in a vectorial way we express an elemental lenght of this filmaent as $\vec{dl}$ . So,

$\vec{dl}x\vec{r} = R*dl*\vec{A}$

Where $\vec{A}$ imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

$\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}$

Substituting the preoviusly equation obtained,

$\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}$

$\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}$

$\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}$

So we can express the velocity induced is,

$\vec{V} = \frac{\Gamma}{2R}\vec{A}$

6 0

3 years ago

What heavy element was the chernobyl nuclear accelarators supposed to make?​

What heavy element was the chernobyl nuclear accelarators supposed to make?