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dybincka [34]
3 years ago
12

Which planet could float on water? Mercury Earth Jupiter Saturn Uranus

Physics
2 answers:
sammy [17]3 years ago
7 0

with enough water, Saturn will float on water as its density is 687kg/m3.

Volgvan3 years ago
5 0

Answer:

Explanation:

saturn is the lightest of all planets in the solar system n is light enough to float on water

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A snail and an inchworm are in a race. Their race track heads north for a distance of 2 m. If the inchworm comes to the end of t
olga2289 [7]
Well the basic equation for velocity is v=d/t where d is distance and t is time. So v=2m/50s and the answer is v=0.04meter/second.
7 0
3 years ago
Help plzzzz thank youu
dybincka [34]
The answer is B I hope this helps luv
5 0
4 years ago
Mercury has a radial acceleration of 3.96 × 10−2 m/s2 and its orbital period is T = 88 days. What is the radius of Mercury’s orb
Maslowich

Answer: 58,045,522,878.8 meters

Explanation:

Ok, the data we have is

Period = T = 88 days

Radial acceleration = ar = 3.96x10^-2 m/s^2

And we know that the equation for the radial acceleration is:

ar = v^2/r = r*w^2

Where v is the velocity. r is the radius and w is the angular velocity.

And we know that:

w = 2*pi*f

where f is the frequency, and:

T = 1/f.

Then we can write:

w = 2*pi/T

and our equation becomes:

ar = r*(2*pi/T)^2

Now we solve this for r.

First we need to use the same units in both equations, so we want to write T in seconds.

T = 88 days,

A day has 24 hours, and one hour has 3600 seconds:

T = 88*24*3600 s =7,603,200s

Then:

3.96x10^-2 m/s^2 = r*(2*3.14/7,603,200s)^2

r = (3.96x10^-2 m/s^2) /(2*3.14/7,603,200s)^2 = 58,045,522,878.8 meters

5 0
3 years ago
Determine the Mutual Inductance per unit length between two long solenoids, one inside the other, whose radii are r1 and r2 (r2
Triss [41]

Answer:

M' = μ₀n₁n₂πr₂²

Explanation:

Since r₂ < r₁ the mutual inductance M = N₂Ф₂₁/i₁ where N₂ = number of turns of solenoid 2 = n₂l where n₂ = number of turns per unit length of solenoid 2 and l = length of solenoid, Ф₂₁ = flux in solenoid 2 due to magnetic field in solenoid 1 = B₁A₂ where B₁ = magnetic field due to solenoid 1 = μ₀n₁i₁ where μ₀ = permeability of free space, n₁ = number of turns per unit length of solenoid 1 and i₁ = current in solenoid 1. A₂ = area of solenoid 2 = πr₂² where r₂ = radius of solenoid 2.

So, M = N₂Ф₂₁/i₁

substituting the values of the variables into the equation, we have

M = N₂Ф₂₁/i₁

M = N₂B₁A₂/i₁

M = n₂lμ₀n₁i₁πr₂²/i₁

M = lμ₀n₁n₂πr₂²

So, the mutual inductance per unit length is M' = M/l = μ₀n₁n₂πr₂²

M' = μ₀n₁n₂πr₂²

3 0
3 years ago
Is O2 considered one atom or 2 atoms?
natulia [17]
O2 is considered 2 atoms because O2 is 2 oxygen atoms.

4 0
4 years ago
Read 2 more answers
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