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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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The distribution circuit of a residential power line is operated at 2000 V rms. This voltage must be reduced to 240 V rms for us
forsale [732]

Answer:

N_p\approx3958

Explanation:

In an ideal transformer the relationship between the voltages is proportional to the ratio between the number of turns of the windings. Thus:

\frac{V_p}{V_s} =\frac{N_p}{N_s}

Where:

V_p=Voltage\hspace{3}in\hspace{3}primary\hspace{3}coil

V_s=Voltage\hspace{3}in\hspace{3}secondary\hspace{3}coil

N_p=Turns\hspace{3}on\hspace{3}primary\hspace{3}coil

N_s=Turns\hspace{3}on\hspace{3}secondary\hspace{3}coil

So, solving for N_p

N_p=N_s*\frac{V_p}{V_s} =475*\frac{2000}{240} =3958.333333\approx3958

7 0
3 years ago
If two teams playing tug-of-war pull on a rope with equal but opposite forces, what is the net external force on the rope?
Rashid [163]
0, Because if they are pulling with equal force then they cancel each other out<span />
3 0
3 years ago
In a circus performance, a monkey on a sled is given an initial speed of 4.3 m/s up a 24◦ incline. The combined mass of the monk
Serggg [28]

Answer:

d = 1.15 m

Explanation:

  • In absence of friction, the change in kinetic energy of the combined mass of the monkey and the sled, must be equal (with opposite sign), to the change in gravitational potential energy:

        ΔK = -ΔU

  • When friction is not negligible, the change in mechanical energy, must be equal to the work done by non-conservative forces (kinetic friction in this case):

       ΔK + ΔU = Wnc (1)

  • As the monkey + sled reach to the maximum distance up the incline, they will come momentarily to a stop, so the final kinetic energy is 0.

        (K_{f} -K_{o}) = 0 - \frac{1}{2} * m*v_{o} ^{2} = -\frac{1}{2} *22.5kg*(4.3m/s)^{2} = -208.1J

  • The change in gravitational energy, can be written as follows:

        (U_{f} - U_{o} ) = m*g*h - 0 = m*g*h = \\ \\ 22.5 kg*9.8 m/s2*d*sin (24 deg) = 89.7J*d

  • The sum of these two quantities, must be equal to the work done by the friction force, along the distance d up the incline:

        W_{nc} = -\mu k*N*d

  • The normal force, always normal to the surface, must be equal and opposite to the component of the weight normal to the incline:

        N = m*g*cos \theta = m*g*cos (24 deg) = \\ \\ 22.5 kg*9.8m/s2*0.913 = 201.4 N

  • Replacing in the equation for Wnc:

        W_{nc} = -\mu k*N*d = -0.45*201.4 N*d = -90.6 N*d

  • We can return to the equation (1) and solve for d:

        -208.1 J + 89.7N*d = -90.6N*d\\\\  d = \frac{208.1}{180.3} =1.15 m

3 0
3 years ago
An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha
Lynna [10]

Answer: 6.45 s

Explanation:

We have the following equation:

y=y_{o}+V_{o}t-\frac{1}{2}gt^{2} (1)

Where:

y=0 is the height when the rock hits the ground

y_{o}=75 m the height at the edge of the cilff

V_{o}=20 m/s the initial velocity

g=9.8 m/s^{2} acceleration due gravity

t time

0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}  (2)

Rearranging the equation:

-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0 (3)

At this point we have a quadratic equation of the form at^{2}+bt+c=0, and we have to use the quadratic formula if we want to find  t:

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}  (4)

Where a=-4.9, b=20, c=75

Substituting the known values and choosing the positive result of the equation:

t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}  (5)

t=6.453 s  This is the time it takes to the rock to hit the ground

8 0
3 years ago
Calculate the number of molecules of hydrogen and carbon present in 4 g of methane?​
emmasim [6.3K]

Answer: 6.022×10²³ molecules of CH4 which consists of 1 mole of C atoms and 4 moles of H atoms.

4 0
3 years ago
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