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1 year ago

3. a) If the distance between two objects increases by a factor of 5 (5),

Physics

1 answer:

irina [24]1 year ago

3 0

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

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Can yall please help me with this?

Lostsunrise [7]

Answer: b

Explanation:

B is a chemical property

Reference what are physical properties

And what are chemical properties

Under certain conditions it can release a gas is the key words letting you know this is a chemical property

8 0

2 years ago

Which of the following is NOT a type of acceleration?

otez555 [7]

It’s C because just trust

5 0

2 years ago

Convert 0.700 atm of pressure to its equivalent in millimeters of mercury.

inna [77]

Answer:

532 millimeters of mercury

Explanation:

In order to convert the pressure from atm to millimeters of mercury (mm Hg), we should remind the conversion factor between the two units:

1 atm = 760 mm Hg

Therefore, we can solve the problem by setting up the following proportion:

$1 atm : 760 mmHg = 0.700 atm : x$

Solving for x, we find

$x=\frac{(760 mmHg)(0.700 atm)}{1 atm}=532 mmHg$

5 0

3 years ago

At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex

telo118 [61]

Answer:

346 * 10⁶ m

Explanation:

The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let $F_{e}$ be the force of gravity exerted by the earth

and let $F_{m}$ be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d, is given by:

$F = \frac{Gm_{1} m_{2} }{d^{2} }$

Mass of the earth, $m_{e} = 5.97 * 10^{24} kg$

Mass of the moon, $m_{m} = 7.348 * 10^{22} kg$

Mass of the satellite, $m_{s} = ?$

$F_{e} = \frac{G*5.97 * 10^{24} M }{d^{2} }$ ...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

$F_{m} = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }$ ............................(2)

Equating equations (1) and (2)

$\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }$

$(5.97 * 10^{24})(14.78 * 10^{16} -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d + 5.97 * 10^{24}d^{2} = 7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\$