4. A car accelerating at 60 meters/second is hit in an accident by a bus. The net force exerted on
1 answer:
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Answer:
0.83 g/cm³
Explanation:
The volume of the alcohol is ...
(33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³
The density of water is about 1 g/cm³, so the volume of 9 g of it is ...
(9 g)/(1 g/cm³) = 9 cm³
The total volume is ...
42.0253 cm³ +9 cm³ = 51.0253 cm³
The total mass is ...
33.2 g + 9 g = 42.4 g
So, the resulting density is ...
(42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³
The resulting mixture has a density of about 0.83 g/cm³.
_____
<em>Additional comment</em>
Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.
The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...
(33.2/46)/(33.2/46 +9/18.02) ≈ 0.59
The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.
We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV
Answer:
the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Explanation:
Mass of satellite, m
orbit radius of first, r1 = r
orbit radius of second, r2 = 2r
Centripetal force is given by
Where v be the orbital velocity, which is given by
So, the centripetal force is given by
where, g bet the acceleration due to gravity
So, the centripetal force
Gravitational force on the satellite having larger orbit
.... (1)
Gravitational force on the satellite having smaller orbit
.... (2)
Comparing (1) and (2),
F' = 4 F
So, the centripetal force on the satellite in the larger orbit is _one fourth_ as that on the satellite in the smaller orbit.
Explanation:
Formula to represent thrust is as follows.
F =
=
or, p =
F =
=
= 201.67 N
Thus, we can conclude that the thrust is 201.67 N.