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sertanlavr [38]
3 years ago
15

What will keep the gravitational force between two objects unchanged?

Physics
1 answer:
Luden [163]3 years ago
6 0

The gravitational force between two objects is determined by

-- the product of their masses, and

-- the distance between their centers.

If you want to keep the gravitational force between two objects unchanged, then you can make changes to the mass of one object, the mass of the other object, and the distance between them, BUT you have to be very careful to always keep the quantity

(one mass) x (the other mass) / (the distance)²

the same, unchanged number.  This means that if you change one thing, you'll have to change something else to keep that expression constant.

For example:  If one of the masses suddenly somehow becomes 4 times as great, then you can EITHER change the other mass to 1/4 of its original mass, OR move the two objects twice as far apart.

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To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1
dedylja [7]

Answer:

2.2\times 10^{-5} s

Explanation:

We are given that  

The wavelength of sound wave=\lambda=3.3 cm=3.3\times 10^{-2}m/s

1 cm/s=10^{-2}m/s

Speed of sound wave,v=1522 m/s

We have to find the period of the wave.

We know that

Frequency=\nu=\frac{v}{\lambda}

Using the formula

Frequency =\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4} Hz

Time period=\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s

Using identity:\frac{a^x}{a^y}=a^{x-y}

Hence, the time period of the wave=2.2\times 10^{-5} s

4 0
3 years ago
A spring that is stretched by hanging a 5kg mass. It's equilibrium length was 0.5 meters. Now the length of the spring is 1.6 me
gtnhenbr [62]
45 N/m as attached, please comment

5 0
3 years ago
Because of changes over time, the most accurate weather forecasts are: A) analog forecasts. B) long-term forecasts. C) seven-day
blondinia [14]
The answer is analog forecasts
8 0
2 years ago
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You make tea with 0.50 kg of 85.0°C water and let it cool to room temperature 120.0°C2.
dedylja [7]

Answer:

Explanation:

a ) Entropy change dS = dQ/T

= mcdT /T

Integrating both sides

S₂ - S₁ = - mclnT₂ /T₁

= - .5 X 4200 ln (85+273) /( 20 + 273 )

.5 X 4200 ln 358/ 293

= - 417.6 J/K

Entropy change will be negative as heat is lost by the system .

b ) Sine there is no change in the temperature of air , This heat will enter air at temperature ( 20+ 273) K = 293 K

Heat entering air

= .5 x 4200 x 65

= 136500 J

Change in entropy

136500 / 293 ( room temperature is constant at 293k

= + 465.87 J/K

Entropy change will be positive  as heat is gained  by the system .

Total change in the entropy of the system (tea + air )

= +465.87 - 417.6

= 48.27 J/K

Entropy change will be negative as heat is lost by the system .

5 0
3 years ago
What is necessary for a convection cell to be set up in a fluid?
Vitek1552 [10]

Convection cells don't work in solids.

In order for convection to happen, material with some temperature and density has to FLOW to a place where the material there has a different temperature and density.

Material doesn't FLOW in solids, only in FLuids.

3 0
3 years ago
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