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scZoUnD [109]
3 years ago
7

Calculate the force between two objects that have masses of 70 Kilograms and 2,000 kilograms separated by a distance of 1 meter

Physics
1 answer:
Hunter-Best [27]3 years ago
5 0

9.3 x 10⁻⁶N

Explanation:

Given parameters:

Mass 1 = 70kg

Mass 2 = 2000kg

distance = 1m

Unknown:

force between them =  

Solution:

The force between the two masses will be a gravitational force of attraction.

  F = \frac{G m_{1}m_{2}  }{r^{2} }

 G is universal gravitation constant = 6.67430×10−¹¹ N⋅m²/kg²

 r is the distance between the two masses

Substituting the parameters:

 F = \frac{6.67 x 10^{-11} x 70 x 2000}{1^{2} } = 9.3 x 10⁻⁶N

 Learn more:

Universal gravitation constant   brainly.com/question/1724648

#learnwithBrainly

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Answer:

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3 years ago
A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s2. Secret agent Austin Powers jumps on ju
denpristay [2]

Answer:

a) h=250\ m

b) \Delta h=0.0835\ m

Explanation:

Given:

  • upward acceleration of the helicopter, a=5\ m.s^{-2}
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a)

<u>Maximum height reached by the helicopter:</u>

using the equation of motion,

h=u.t+\frac{1}{2} a.t^2

where:

u = initial velocity of the helicopter = 0 (took-off from ground)

t = time of observation

h=0+0.5\times 5\times 10^2

h=250\ m

b)

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<u>height fallen freely by Austin:</u>

h_f=u.t_f+\frac{1}{2} g.t_f^2

where:

u= initial velocity of fall at the top = 0 (begins from the max height where the system is momentarily at rest)

t_f= time of free fall

h_f=0+0.5\times 9.8\times 7^2

h_f=240.1\ m

<u>Velocity just before opening the parachute:</u>

v_f=u+g.t_f

v_f=0+9.8\times 7

v_f=68.6\ m.s^{-1}

<u>Time taken by the helicopter to fall:</u>

h=u.t_h+\frac{1}{2} g.t_h^2

where:

u= initial velocity of the helicopter just before it begins falling freely = 0

t_h= time taken by the helicopter to fall on ground

h= height from where it falls = 250 m

now,

250=0+0.5\times 9.8\times t_h^2

t_h=7.1429\ s

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<u>remaining time,</u>

t'=t_h-t_f

t'=7.1428-7

t'=0.1428\ s

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h'=v_f.t'+\frac{1}{2} a_p.t'^2

h'=68.6\times 0.1428+0.5\times 2\times 0.1428^2

h'=9.8165\ m

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\Delta h=h-(h_f+h')

\Delta h=250-(240.1+9.8165)

\Delta h=0.0835\ m

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strojnjashka [21]

Answer:

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