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aniked [119]
3 years ago
8

The acceleration due to gravity on Jupiter is 2.5 times what it is here on earth. An object weighing 347.9 N here on earth will

have what mass and weight on Jupiter?
Physics
1 answer:
inessss [21]3 years ago
5 0

Answer:  weight on Jupiter = 869.75 N

              mass  on Earth = mass on Jupiter = 35.5 Kg

Explanation:

W = mg

W = weight

m = mass

g = gravitational acceleration [ on the Earth, g₁ = 9,8 N/kg ]

On the Earth,

G₁ = m x g₁  = 347,9 N

On the Jupiter,

G₂ = mg₂  

mass on the Earth = mass on the Jupiter !  

m = G₁ : g = 347.9 N : 9,8 N/kg = 35.5 kg

G2 : G1 = 2.5

G₂ = 2,5 G₁ = 2,5 x 347.9 N =  869,75 N

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What is the separation distance, in meters, between masses m1 = 15 x 107 kg and m2 = 62 x 107 kg when the gravitational force be
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Answer:

94.1 m

Explanation:

From Coulombs law,

F  = Gm1m2/r²................... Equation 1

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Make r the subject of the equation,

r = √(Gm1m2/F)................. Equation 2

Given: F = 7×10² N, m1 = 15×10⁷ kg, m2 = 62×10⁷ kg,

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Substitute into equation 2

r = √( 6.67×10⁻¹¹×15×10⁷×62×10⁷/7×10²)

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If a bean of mass 2.0g jumps 1.0cm from your hand into the air, how much potential energy has it gained in reaching its highest
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If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is u  =  7.73 \  m/s

Explanation:

From the question we are told that

   The height at which he let go of the brief case is  h =  130 m  

    The  time taken before the the brief case hits the water is  t =  6 s

Generally the initial speed of the  briefcase (Which also the speed of the helicopter )before the man let go of it is  mathematically evaluated using kinematic equation as

      s = h+  u t +  0.5 gt^2

Here s  is the distance covered by the bag at sea level which is zero

      0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>    0 = 130+  u * (6) +  0.5  *  (-9.8) * (6)^2

=>   u  =  \frac{-130 +  (0.5 * 9.8 *  6^2) }{6}

=>   u  =  7.73 \  m/s

     

7 0
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