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neonofarm [45]
3 years ago
6

Will is a scientist. He’s designing a spacecraft that would allow people to land on Mars. Will’s mass on Earth is 75 kilograms.

Will knows that the gravitational pull of Mars is less than the gravitational pull of Earth. When he lands on Mars, his mass will be (Less than, Equal to or More than) 75 kilograms. His weight on Mars will be (less than, Equal to, or more than) his weight on Earth.
Physics
2 answers:
s2008m [1.1K]3 years ago
8 0

Mass is the amount of substance present inside the body.

As per Einstein's relativity the mass of the body increases with the increase of the speed of the body.

Mathematically it is given as m=\frac{m_{0} }{\sqrt{1-\frac{v^2}{c^2} } }

where m is the mass of the body during motion and m_{0} is the rest mass.

Here V is the velocity of the body and c is the speed of light.

But during non-relativistic situation i.e the velocity of the body is very very less as compared to the velocity of the light or equal to zero, the mass of the substance remains unchanged.

Hence mass of Will will be the same both in Mars and Earth i.e 75 kg

Again the weight of a body is mathematically given as -

                             weight= mass×acceleration due to gravity

                           i.e W= m×g

The value of g is less in Mars as compared to Earth.

Hence the weight of Will on Mars is less as compared to weight on the Earth.

Vesnalui [34]3 years ago
6 0
Mass will remain unchanged, always.  His weight, which is the gravitational force acting on that mass will be less in this case.
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Answer:

The kinetic energy lost in the collision is 48 J

Explanation:

Given;

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mass of the second ball, m₂ = 6.0 kg

initial speed of the first ball, u₁ = 12 m/s

initial speed of the second ball, u₂ = 4 m/s

let v be the final velocity of the two balls after the inelastic collision

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = v(m₁ + m₂)

2 x 12 + 6 x 4 = v(2 + 6)

48 =  8v

48 / 8 = v

v = 6 m/s

The initial kinetic energy of the balls is calculated as;

K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²

K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²

K.E₁ = 144 + 48

K.E₁ = 192 J

The final kinetic of the balls is calculated as;

K.E₂ = ¹/₂(m₁ + m₂)(v²)

K.E₂ = ¹/₂(2 + 6)(6²)

K.E₂ = ¹/₂(8)(6²)

K.E₂ = 144 J

The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J

Therefore, the kinetic energy lost in the collision is 48 J

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