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3 years ago

How to solve number 10

Physics

1 answer:

Vinil7 [7]3 years ago

6 0

The 'formulas' to use are just the definitions of 'power' and 'work':

Power = (work done) / (time to do the work)

and

Work = (force) x (distance) .

Combine these into one. Take the definition of 'Work', and write it in place of 'work' in the definition of power.

Power = (force x distance) / (time)

From the sheet, we know the power, the distance, and the time. So we can use this one formula to find the force.

Power = (force x distance) / (time)

Multiply each side by (time): (Power) x (time) = (force) x (distance)

Divide each side by (distance): Force = (power x time) / (distance).

Look how neat, clean, and simple that is !

Force = (13.3 watts) x (3 seconds) / (4 meters)

Force = (13.3 x 3 / 4) (watt-seconds / meter)

Force = 39.9/4 (joules/meter)

Force = 9.975 Newtons

Is that awesome or what !

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Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

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We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$