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lora16 [44]
3 years ago
10

Un cuerpo de 480 g de masa es atraído con una fuerza de 3.9 E-6 N por otro cuerpo de 196 g de masa. Calcula la distancia a la qu

e se encuentran.
Physics
1 answer:
Soloha48 [4]3 years ago
5 0

Answer:

La distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

Explanation:

Asumamos que ambos cuerpos son partículas, la fuerza de atracción (F), en newtons, entre ambos cuerpos se define mediante la Ley de Newton de la Gravitación Universal, cuya ecuación es:

F = G \cdot \frac{m_{1}\cdot m_{2}}{r^{2}} (1)

Donde:

G - Constante de la gravitación universal, en metros cúbicos por kilogramo-segundo cuadrado.

m_{1}, m_{2} - Masas de los cuerpos, en kilogramos.

r - Distancia entre los cuerpos, en metros.

Si sabemos que G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}, m_{1} = 0.48\,kg, m_{2} = 0.196\,kg y F = 3.9\times 10^{-6}\,N, entonces la distancia entre los dos cuerpos es:

r = \sqrt{\frac{G\cdot m_{1}\cdot m_{2}}{F} }

r \approx 1.269\times 10^{-3}\,m

Es decir, la distancia entre los dos cuerpos es aproximadamente 1.269 milímetros.

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Answer:

1. S-S Repulsion           N-S Attraction

2. S-N Attraction          N-N Repulsion

Explanation:

1. S-S Repulsion           N-S Attraction

2. S-N Attraction          N-N Repulsion

3 0
3 years ago
Two Charged spheres of 6.4 nC charge are 16 cm apart. What is the magnitude of the field in middle? ( n = 10^-9, 100 cm = 1 m)
emmainna [20.7K]

Answer:

zero

Explanation:

q = 6.4 nC = 6.4 x 1 0^-9 C

d = 16 cm = 0.16 m

r = 16 / 2 = 8 cm = 0.08 m

Electric field at P due to the charge placed at A

Ea = k q / r^2

Ea = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards right

Ea = 9000 Towards right

Electric field at P due to the charge placed at B

Eb = k q / r^2

Eb = ( 9 x 10^9 x 6.4 x 10^-9) / (0.08 x 0.08) Towards left

Eb = 9000 Towards left

The magnitude of electric field is same but teh direction is opposite, so the resultant electric field at P is zero.

3 0
3 years ago
Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am
Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg

So, from friction law

Fr=µN

For static

Fs=µsN

For kinetic or dynamic

Fk= µkN

Using newton law

Along x-axis

Before the body start moving we can get the Force and since the force is the same use to start the block in motion.

Then,

ΣF = ma

Since at static the body is not moving then, a=0

F-Fs=0

F=Fs

Since, Fs=µsN

F=Fs=µsN

Then, the force to keep the body in motion too is F=µsN

Now analyses when the body is in motion

ΣF = ma

F-Fk=ma

ma=F - Fk

Substituting F=µsN and Fk=µkN

ma=µsN - µkN

ma=N(µs - µk)

Since N=mg

Then, ma=mg(µs - µk)

m cancels out, then

a=g(µs - µk)

Then the acceleration of the body is given as "a=g(µs - µk)"

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