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Zielflug [23.3K]
2 years ago
15

The gravitational force produce between any two object kept 2.5×10 to the power 4 km apart is 580N.At what distance should they

kept so that the grqvitation force becomes half.​
Physics
1 answer:
timofeeve [1]2 years ago
3 0

Answer:

d = 3.54 x 10⁴ Km

Explanation:

Given,

The distance between the two objects, r = 2.5 x 10⁴  Km

The gravitational force between them, F = 580 N

The gravitational force between the two objects is given by the formula

                                         F = GMm/r² newton

When the gravitational force becomes half, then the distance between them becomes

Let us multiply the above equation by 1/2 on both sides

                                        ( 1/2) F = (1/2) GMm/r²

                                                   =  GMm/2r²

                                                   =  GMm/(√2r)²

Therefore, the distance becomes √2d, when the gravitational force between them becomes half

                                           d = √2r = √2 x 2.5 x 10⁴  Km

                                               = 3.54 x 10⁴  Km

Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴  Km so that the gravitational force becomes half.

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Answer

given,

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so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

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2 years ago
A large container, 120 cm deep is filled with water. If a small hole is punched in its side 77.0 cm from the top, at what initia
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The water will flow at a speed of 3,884 m/s

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Torricelli's equation

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v = \sqrt{2*9,8m/s^2*0,77m} = 3,884 m/s

6 0
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A woman fires a rifle with barrel length of 0.5400 m. Let (0, 0) be where the 125 g bullet begins to move, and the bullet travel
dybincka [34]

Answer:

Explanation:

Given that,

Length of barrel =0.54m

Mass of bullet=125g=0.125kg

Force extend

F=16,000+10,000x-26,000x²

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W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=0.54m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=0.54m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=0.54m

W= 16,000(0.54) + 5000(0.54²) - 8666.667(0.54³) +0+0-0

W=8640+1458-1364.69

W=8733.31J

The workdone by the gas on the bullet is 8733.31J

b. Work done is given as

Work done when the length=1.05m

W= ∫Fdx

W= ∫(16,000+10,000x-26,000x² dx from x=0 to x=1.05m

W=16,000x+10,000x²/2 -26,000x³/3 from x=0 to x=1.05m

W=16,000x+5,000x²- 8666.667x³ from x=0 to x=1.05mm

W= 16,000(1.05) + 5000(1.05²) - 8666.667(1.05³) +0+0-0

W=16800+5512.5-10032.75

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The workdone by the gas on the bullet is 12,279.75J

3 0
2 years ago
Imagine you derive the following expression by analyzing the physics of a particular system: M= (mv2r)(mGr2). Simplify the expre
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Answer:

The simplified expression is M  =  \frac{v^2 r}{G}

Explanation:

From the question we are told that  

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So simplifying we have

    M  =    \frac{m v^2}{r} *  \frac{r^2 }{ mG }

    M  =  \frac{v^2 r}{G}

Thus the simplified formula is M  =  \frac{v^2 r}{G}

3 0
2 years ago
The magnetic field around a current-carrying wire is ________proportional to the current and _________proportional to the distan
PSYCHO15rus [73]

Answer:Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

Explanation:

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           r= distance from the current carrying wire

Thus, The magnetic field around a current-carrying wire is <u><em>directly</em></u>  proportional to the current and <u><em>inversely</em></u>  proportional to the distance from the wire.  

Now if I'=3I and r'=2r then magnetic field B' is given by

B'=\frac{\mu _oI'}{2\pi r'}=\frac{\mu _o3I}{2\pi 2r}=1.5B

Thus If the current triples while the distance doubles, the strength of the magnetic field increases by <u><em>one and half (1.5)</em></u> times.

   

7 0
2 years ago
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