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3 years ago

Derive a relation between Universal Gravitational Constant (G) and Acceleration due to gravity(g)?

2 answers:

5 0

suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

snow_tiger [21]3 years ago

4 0

Suppose earth is a soid sphere which will attract the body towards its centre.So, acc. to law of gravitation force on the body will be,
F=G*m1m2/R^2
but we now that F=ma
and here accleration(a)=accleration due to gravity(g),so
force applied by earth on will also be mg
replace above F in formula by mg and solve,
F=G*mE*m/R^2 ( here mE is mass of earth and m is mass of body)
mg=G*mEm/R^2
so,
g =G*mE/R^2

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A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume

Assoli18 [71]

Answer:

Hence the pressure is $3\times 10^5 Pa$

Explanation:

Given data

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3 years ago

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Answer:

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2 years ago

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Ron is on a Ferris wheel of radius 30 ft that turns counterclockwise at a rate of one revolution every 12 seconds. The lowest po

alexgriva [62]

Answer:

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

Explanation:

1 full revolution is $2\pi.$ let \theta be the angle of Ron's position.

At t = 0. $\theta = 0$

one full revolution occurs in 12 sec, so his angle at t time is

$\theta =2\pi \frac{t}{12} = \frac{\pi}{6}t$

r is radius of circle and it is given as

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$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t$

however, that is centered at (0,0) and the positioned at time t = 0 is (30,0). it is need to shift so that the start position is (30,45). it can be done by adding to y

$x = 30cos\frac{\pi}{6}t$

$y = 30sin\frac{\pi}{6}t + 45$

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