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3 years ago

7

the gravational force is between two objects 18 N . if the distance between them is tripled and their masses are doubled. then w

hat is the gravitational force of the two objects

2 answers:

Anni [7]3 years ago

8 0

$F_1 = \frac{G m_1 m_2 }{r^2 } = 18N

F_2 = \frac{G M_1 M_2 }{R^2}

R = 3r

M_1 = 2m_1

M_2 = 2m_2

F_2 = \frac{G 2m_1 2m_2}{(3r)^2} = \frac{4}{9} * \frac{G m_1m_2}{r^2} = \frac{4}{9} * 18N

$

Alla [95]3 years ago

8 0

Answer:

It would be

C. or 18 N

∫∫- Depree

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Nadya [2.5K]

1) The mass of the continent is $3.3\cdot 10^{21}kg$

2) The kinetic energy of the continent is 1683 J

3) The speed of the jogger must be 6.57 m/s

Explanation:

1)

The continent can be represented as a slab of size

$d=5850 km = 5.85\cdot 10^6 m$

and depth

$t = 35 km = 3.5\cdot 10^4 m$

So its volume is

$V=d^2 t = (5.85\cdot 10^6)^2(3.5\cdot 10^4)=1.20\cdot 10^{18} m^3$

We also know that the density of the continent is

$\rho = 2750 kg/m^3$

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$m=\rho V=(2750)(1.20\cdot 10^{18})=3.3\cdot 10^{21} kg$

2)

The kinetic energy of the continent is given by

$K=\frac{1}{2}mv^2$

where

m is its mass

v is its speed

We have already calculate its mass, while the speed is

v = 3.2 cm/year

We have to convert into SI units first, as follows:

$v=3.2 \frac{cm}{year} \cdot \frac{1}{100 cm/m} \cdot \frac{1}{(365 d/y)(24h/d)(60min/h)(60 s/min)}=1.01\cdot 10^{-9} m/s$

The mass is

$m=3.3\cdot 10^{21} kg$

So, the kinetic energy of the continent is

$K=\frac{1}{2}(3.3\cdot 10^{21})(1.01\cdot 10^{-9})^2=1683 J$

3)

Here we have a jogger having the same kinetic energy of the continent, so

$K=1683 J$

And the kinetic energy of the jogger can be expressed as

$K=\frac{1}{2}mv^2$

where

m = 78 kg is the mass of the jogger

v is his speed

We can therefore re-arrange the equation to find the speed of the man, and we get:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1683)}{78}}=6.57 m/s$

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3 years ago

1. A boy walk 140 m due north, 85 m due east, 35 m due southeast, 38 m due west and then 19 m due northwest. Calculate the displ

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Answer:

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Explanation:

Let E be along the positive x axis of a unit circle

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north displacement

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In the first trial of this problem, a net force of $F_{net}$ is applied to the object, causing an acceleration of

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