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Fofino [41]
3 years ago
6

¿Cómo obtener razonadamente la primera ley de newton a partir de la segunda?

Physics
1 answer:
storchak [24]3 years ago
5 0
Do you speak English? If so I can help you in the comments
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Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula
velikii [3]

Answer:

6.1328 kg

60.16284 N

Explanation:

r = Radius of ball = 0.11 m

\rho = Density of fluid = 1.1\times 10^3\ kg/m^3 (Assumed)

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of ball

V = Volume of ball = \frac{4}{3}\pi r^3

The weight of the bowling ball will balance the buouyant force

W=F_b\\\Rightarrow mg=V\rho g\\\Rightarrow m=\frac{V\rho g}{g}\\\Rightarrow m=V\rho\\\Rightarrow m=\frac{4}{3}\pi 0.11^3\times 1.1\times 10^3\\\Rightarrow m=6.1328\ kg

The mass of the bowling ball will be 6.1328 kg

Weight will be 6.1328\times 9.81=60.16284\ N

5 0
3 years ago
Why are the orbits of planets only nearly circular and not perfectly circular?
Galina-37 [17]

Explanation:

this is my answer this is helpful for you

7 0
3 years ago
Describe three physical changes that occur in nature?
Alex Ar [27]

Answer: Physical changes in nature could then be erosion in a mountain, the melting of snow, and a river freezing over from the cold. Since none of these changes affect the chemical composition of the mountain, the snow, or the river, they are physical changes.

Explanation:

8 0
3 years ago
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As you can see, my cousin has a lot of hair. He uses an 1800 W blow dryer and it takes him
maw [93]

Power = 1800W (or 1.8KW by dividing by 1000)

Time = 3 hours

Power = energy/ time

1.8KW = energy/ 3

x3

5.4Kw/h= energy

(5.4KJ or 5400J used)

$0.15 Kw/h

$0.15 X 5.4 = 0.81

Thus, cost $0.81

Hope this helps!

5 0
2 years ago
a tire with inner volume of 0.0250m^3 is filled with air at a gauge pressure of 36.0 psi. If the tire valve is opened to the atm
enyata [817]

Answer: Escaped volume = 0.0612m^3

Explanation:

According to Boyle's law

P1V1 = P2V2

P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)

P2 = atmospheric pressure= 14.696psi

V1 = volume of tire =0.025m^3

V2 = escaped volume + V1 ( since air still remain in the tire)

V2 = P1V1/P2

V2 = 50.696×0.025/14.696

V2 = 0.0862m^3

Escaped volume = 0.0862 - 0.025 = 0.0612m^3

5 0
3 years ago
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