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3 years ago

What would be the weight of a 59.1-kg astronaut on a planet with the same density as Earth and having twice Earth's radius?

1 answer:

5 0

You could always directly use F = G M m / r², but that takes forever.

<span>It is best to know that, on Earth's surface, the astronaut would weigh </span>
<span>F = m g = 579.2 N </span>
<span>where m = 59.1 kg and g = 9.8 m/s². </span>

<span>You then notice from F = G M m / r² that doubling the mass of the earth would double his weight and doubling the radius would give 1/4 his original weight. So... </span>
<span>F = 1/4 * 2 * 579.2 N = 290 N </span>

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What Do You Already Know about Density? Material Design. Number each material and sort the items in order from lowest (1) to hig

Elena-2011 [213]

Answer:

1. Dry Beans - 591.75 kg/m^3

2. Flour - 593 kg/m^3

3. Wax - 900 kg/m^3

4. Wet sand - 2039 kg/m^3

5. Chalk - 2499 kg/m^3

6. Talcum Powder - 2776 kg/m^3

7. Copper - 8960 kg/m^3

Explanation:

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3 years ago

A 230.-mL sample of a 0.240 M solution is left on a hot plate overnight; the following morning, the solution is 1.75 M. What vol

Gwar [14]

Answer:

The volume of water evaporated is 199mL

Explanation:

Concentration is calculated with the following formula

$C=\frac{n}{V}$

where n is the number of moles of solute and V is the volume of the solution (in this case is the same as the solvent volume) in liters.

So we isolate the variable n to know the amount of moles, using the volume given in liters

$230mL=0.23L$

$n=C*V=0.240 M*0.23L=0.055 mol$

Now, we isolate the variable V to know the new volume with the new concentration given.

$V=\frac{n}{C} =0.055mol/1.75M=0.031L=31mL$

Finally, the volume of water evaporated is the difference between initial and final volume.

$V_{ev}= V_{i} -V_{f} =230mL-31mL=199mL$

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3 years ago

Which of the following is the best explanation of work

Hunter-Best [27]

Answer:

Explanation:

5 0

3 years ago

Read 2 more answers

S/REF No. Date If the load distance of a level is 20 cm and effort distance is 6ocm, calculate the amount of effort required to

andrew11 [14]

The answer is 602 cause I added

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3 years ago

A brick lands 10.1 m from the base of a building. If it was given an initial velocity of 8.6 m/s [61º above the horizontal], how

Montano1993 [528]

<h2>Answer: 10.52m</h2><h2 />

First, we have to establish the <u>reference system</u>. Let's assume that the building is on the negative y-axis and that the brick was thrown at the origin (see figure attached).

According to this, the initial velocity $V_{o}$ has two components, because the brick was thrown at an angle $\alpha=61\º$ :