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2 years ago

Calculating Welght What is the weight of a 45 kg box? N

Physics

1 answer:

Julli [10]2 years ago

7 0

Answer:

Im not sure if this is right but here is the answer i think it is....So multiply 45 by 2.205 to get 99.225 pounds. Alternately, to convert from kg to Newtons, use the fact that 1 kg is 9.8 Newtons

Explanation:

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Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p

PSYCHO15rus [73]

Answer:

Energy density will be 14.73 $J/m^3$

Explanation:

We have given capacitance $C=225\mu F=225\times 10^{-6}F$

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm $0.2\times 10^{-3}m$

We know that there is relation between electric field and potential

$E=\frac{V}{d}$ , here E is electric field, V is potential and d is separation between the plates

So $E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C$

Energy density is given by $E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3$

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2 years ago

A variable that is not changed.

AleksAgata [21]

Answer:

A controlled variable does not change during a experiment

Explanation:

it's c

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3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

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3 years ago

What is one of the benefits that scientists have discovered from organisms' daily exposure to radioisotopes?

Sholpan [36]

Answer: 1. Being able to carbon date fossils

From the daily exposure to radioisotopes of organisms, scientists discovered to carbon date fossils. In carbon dating, the age of organic matter can be determined from the relative proportions of C-12 and C-14 i.e. isotopes of Carbon (carbon 12 and carbon 14). This ratio changes over the years because carbon-14 decays and is not replenished as there is no exchange with the atmosphere.

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3 years ago

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HELP ASAP!!

Arada [10]

Percentage error:

1.55 – 1.53 ÷ 1.53
0.02 ÷ 1.53
.013 x 100
1.3 % error

I hope this is right.

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3 years ago

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