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Yanka [14]
3 years ago
13

¿Cuál es la magnitud y dirección de la aceleración de los cuerpos en caída libre en el planeta Tierra?

Physics
1 answer:
Tatiana [17]3 years ago
6 0

La magnitud es de 9.8 m/s² ... la aceleración de la gravedad en o cerca de la superficie de la Tierra.

La dirección es hacia el centro de la Tierra. (Llamamos a esa dirección "abajo").

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Which statement can be made about waves A and B? Wave A has a lower amplitude than Wave B. Wave A has a higher amplitude than Wa
givi [52]

Answer:

Wave A has a higher amplitude than Wave B.

Explanation:

please check the attached image for the waves

5 0
3 years ago
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The speed of a bobsled is increasing because it has an acceleration of 2.0 m/s2. At a given instant in time, the forces resistin
Nookie1986 [14]

Answer:

F = 960 N

Explanation:

given,                      

acceleration = 2 m/s²

Force = 450 N                        

mass = 255 Kg                  

to calculate the force propelling the bobsled forward

∑ F  = ma                                            

F - f = ma                                        

F = m a + f                                  

F = 255 × 2 + 450                          

F = 510 + 450                                

F = 960 N                                    

the magnitude of the force propelling the bobsled forward is F = 960 N

4 0
3 years ago
What happens when sediments are buried ,compacted,and cemented together
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They create sedimentary rock
7 0
3 years ago
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If the odometer on the car reads 25.0 km at the beginning of a trip and 65.0 km a half hour later, what is the average speed of
Y_Kistochka [10]
Distance covered= 65-25= 40 Km
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Speed= 40/0.5= 80 km/hr
6 0
3 years ago
"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th
Tamiku [17]

Answer:

The charge of each charge is 3.02\times10^{-19} C

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})

Because the charges we're dealing are identical positive q1=q2, then (1) is:

F_{e}=k\frac{\mid q^2 \mid}{r^{2}}

Using the values the problem give us:

6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}

solving for q:

q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}

q= 3.02\times10^{-19} C

8 0
3 years ago
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