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k0ka [10]
2 years ago
9

an astronaut on earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. if she were instead in

a lunar module parked on the moon where the gravitation force is 1/6 that of earth, the ice in the same soft drink would float:
Physics
1 answer:
nadya68 [22]2 years ago
3 0

The gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

<h3>What is gravitational force?</h3>

All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.

As we know, the gravitational force is given by:

\rm F = \dfrac{Gm_1m_2}{r^2}

Where G is the gravitational constant.

m1 and m2 are masses.

r is the distance between the masses.

Weight of ice cube = weight of soft drink displaced

mg = ρVg

m = ρV

g does not affect astronauts on earth in a lunar module.

Thus, the gravitation force is 1/6 that of earth, and the ice in the same soft drink would float with 9/10 submerged option (B) is correct.

An astronaut on Earth notes that in her soft drink an ice cube floats with 9/10 of its volume submerged. If she were instead in a lunar module parked on the Moon where the gravitation force is 1/6 that of Earth, the ice in the same soft drink would float:A) with more than 9/10 submerged. B) with 9/10 submerged.C) with 6/10 submerged.D) totally submerged.

Learn more about the gravitational force here:

brainly.com/question/12528243

#SPJ1

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Calculate the force between two objects that have masses of 70 kilograms and 2,000 kilograms separated by a distance of 1 meter.
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▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

The Gravitational Force between given objects will be ~

  • 9.34 \times  {10}^{ - 6}  \:  \: N

\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}

We know that ~

\huge\boxed{\mathrm{F = \dfrac{ Gm_1m_2}{ r²}}}

where ~

  • F = gravitational force

  • m_1 = mass of 1st object = 70 kg

  • m_2 = mass of 2nd object = 2000 kg

  • G = gravitational constant = 6.674 × {10}^ {-11}

  • r = distance between the objects = 1 m

Let's calculate the force ~

  • F =  \dfrac{6.674 \times 10 {}^{ - 11} \times 70 \times 2000 }{1 {}^{2} }

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  • 93.436 \times 10 {}^{ - 7}

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On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
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Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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