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1 year ago

A 1-kg mass at the earth's surface weighs about? a. 1 n. b. 5 n. c. 10 n. d. 12 n. e. none of these

Physics

1 answer:

ValentinkaMS [17]1 year ago

3 0

A 1-kg mass at the earth's surface weighs about C. 10N

The third planet from the Sun is the Earth. It is the seventh largest in terms of size and weighs roughly 5.98 1024 kg. The inherent quality of mass is unaffected by the environment of the object or the technique employed to quantify it.

Newton's law of gravitation can be used to estimate the mass of the Earth. This is set to the fundamental formula, which reads: force (F) = mass (m) times acceleration. Gravitational acceleration (G) is equal to 9.8 m/s2, the Earth's radius is 6.37 106 m, and the gravitational constant (G) is 6.673 1011 Nm2/kg2. The Earth has a mass of 5.96 1024 kg after rearranging the equation and entering all the numbers.

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3 years ago

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Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.

Given $\large{I_{t} = 19.5 A}$ and $\large{I_{B} = 12.5 A}$

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$B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}$

and the magnetic field ( $\large{B_{b}}$ ) at 'P' due to the bottom wire

$B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}$

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$\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T$

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3 years ago

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Answer:

The average velocity of the sled is vavg = s/t.

Explanation:

Hi there!

The average velocity is calculated as the traveled distance over time:

vavg = Δx/Δt

Where:

vavg = average velocity.

Δx = traveled distance.

Δt = elapsed time.

We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:

vavg = s/t

Have a nice day!

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