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Alexxx [7]
3 years ago
10

What force is acting on a 2 kg apple falling on the Earth (g=10)

Physics
1 answer:
jolli1 [7]3 years ago
4 0

Answer:

An apple in free fall accelerates toward the Earth with a free fall acceleration, g. The force of the apple on the Earth also causes the Earth to accelerate toward the falling apple. By Newton's Third Law, the force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth. By Newton,s Second law, the force of the Earth on the apple is equal to the mass of the apple times g , the accelerations due to gravity. And, the force of the the apple on the Earth is equal to the mass of the Earth times the acceleration of the Earth toward the apple. In conclusion, the magnitude of the forces are equal, or

F ( apple on the Earth) = F( the Earth on the apple) or

M( mass of the earth) x a( the acceleration of the earth toward the apple) = m(mass of the apple) x g( the acceleration of the apple toward the Earth) or

a = (m/M) g

Explanation:

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Mumz [18]

Answer:

75 m

Explanation:

The distance x for a uniform acceleration is given by the equation:

x = \frac{1}{2}at^{2}+v_ot+x_o

The given values are:

a = 6 m/s², t = 5 s, v₀ = 0, x₀ = 0

5 0
3 years ago
A spring of force constant 285.0 N/m and unstretched length 0.230 m is stretched by two forces, pulling in opposite directions a
Kaylis [27]

Answer: W = 0.3853 J, e = 0.052 m

Explanation: Given that,

K =285.0N/M , L = 0.230m , F = 15N , e = ?

F = Ke

15 = 285 × e

e = 15÷ 285

e =0.052 m

e + L = 0.052 + 0.230

= 0.282m ( spring new length )

Work needed to stretch the spring

W = 1/2ke2

W = 1/2 × 285 x 0.052 × 0.052

W = 0.3853 J

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Whats an astral projection
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Read 2 more answers
A cart of mass m = 0.12 kg moves with a speed v = 0.45 m/s on a frictionless air track and collides with an identical cart that
lina2011 [118]

Answer:

0.006075Joules

Explanation:

The final kinetic energy of the system is expressed as;

KE = 1/2(m1+m2)v²

m1 and m2 are the masses of the two bodies

v is the final velocity of the bodies after collision

get the final velocity using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

0.12(0.45) + 0/12(0) = (0.12+0.12)v

0.054 = 0.24v

v = 0.054/0.24

v = 0.225m/s

Get the final kinetic energy;

KE = 1/2(m1+m2)v

KE = 1/2(0.12+0.12)(0.225)²

KE = 1/2(0.24)(0.050625)

KE = 0.12*0.050625

KE = 0.006075Joules

Hence the final kinetic energy of the system is 0.006075Joules

5 0
3 years ago
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