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andrew-mc [135]
3 years ago
6

A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What

must be true of the wheelbarrow's motion?
Physics
2 answers:
Tamiku [17]3 years ago
7 0

Answer:

its acceleration must be 0.

Explanation:

According the second Newton's law, the net force applied in a object is equal to the mass of the object multiplied by its acceleration:

F_{net} =m.a

In this case, the net force is

F_{net}=5N-5N=0

What makes sense since both forces have the same value and opposite directions.

Since the mass of the object can't be zero, the acceleration is the one that must be 0 in order fulfill the equation of the second newton's law.

Another option is that, since both forces are opposite, the wheelbarrow could be at rest. Though this is correct, this is not always true.

In conclusion, the only  statement that always is true about the wheelbarrow's motion is that its acceleration must be 0.

mezya [45]3 years ago
4 0
Assuming no other forces are acting on the wheelbarrow, it must be stationary.
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Answer:

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Asimismo, la rapidez angular se describe como función de la aceleración centrípeta:

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Ahora se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R}{a_{r}} }

La aceleración experimentada por el planeta es:

a_{r} = G\cdot \frac{M_{sun}}{R^{2}}

Se reemplaza en la ecuación de período:

T = 2\pi \cdot \sqrt{\frac{R^{3}}{G\cdot M_{sun}} }

La distancia del planeta con respecto al sol es finalmente despejada:

R^{3} = G\cdot M_{sun}\cdot \left(\frac{T}{2\pi} \right)^{2}

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Finalmente, se sustituyen las variables y se determina la distancia:

R = \sqrt[3]{\left(6.674\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (1.989\times 10^{30}\,kg)\cdot \left[\frac{(65\,a)\cdot \left(365\,\frac{d}{a} \right)\cdot \left(86400\,\frac{s}{d} \right)}{2\pi} \right]^{2}}

R \approx 2.418\times 10^{12}\,m

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