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yawa3891 [41]
4 years ago
8

Tres móviles, A, B y C, con las mismas características tienen los siguientes estados: A se encuentra inicialmente en reposo y ca

e libremente, B se mueve hacia la derecha con una velocidad constante de 10m/s y C está en el suelo completamente en reposo
¿Algunos de los tres móviles tendría algún tipo de fuerza que actué sobre él? Si es afirmativa tu respuesta ¿Es o son las únicas fuerzas? Explica.
Physics
1 answer:
Morgarella [4.7K]4 years ago
6 0

Answer:

La respuesta es sí, hay una fuerza que actúa sobre el móvil A y es la única fuerza ya que A cae libremente bajo la influencia de la fuerza.

Explanation:

Según la primera ley de movimiento de Newton, un cuerpo continuará en su estado de reposo o en un movimiento uniforme en línea recta a menos que actúen sobre él fuerzas impresas.

Dado que el móvil A cae libremente, desde su estado de reposo inicial, según la primera ley de movimiento de Newton, experimenta una fuerza que actúa sobre él para hacer que caiga y continúe en caída libre.

El móvil B se mueve con una velocidad constante, por lo tanto, de acuerdo con la primera ley de movimiento de Newton, no hay fuerzas impresas que actúen sobre él.

El móvil C está completamente en reposo en el suelo, por lo tanto, tampoco hay fuerzas que actúen sobre él.

La respuesta es sí, hay una fuerza actuando sobre el móvil A y es la única fuerza cuando A cae libremente bajo la influencia de la fuerza.

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Because polymers are covalently bonded material, and metals are metalicly bonded material. 

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4 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m.how much work
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5 0
3 years ago
How long will the ball be in the air if the cliff is 120 m tall and the ball falls to the base of the cliff?
Readme [11.4K]

Answer:

Explanation:

4.95s≈5s

Use equation :

h=G*t²/2

h=120m ----hight of cliff

G=9.81m/s²

t=?

-----------------------

h=G*t²/2

120m=9.81m/s²*t²/2

240m=9.81m/s²*t²

t²=240m/ 9.81m/s²

t²=24.46s

t=√24.46s²

t=4.95 s≈5s

3 0
4 years ago
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

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ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
4 years ago
Multiply the following and write your answer using significant figures. What is 25 x 15?
Troyanec [42]
25x15 is 375 cndnmekcivjfndn(sorry it said I needed 20 characters to comment)
3 0
3 years ago
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