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Liono4ka [1.6K]
3 years ago
5

La mascota de felipe esta jugando en el parque despues de un tiempo esta agotado de tanto correr. Felipe conoce que el perro con

sumio una porcion de concentrado con la capacidad suficiente de generar 176J. El perro incialmente va a una velocidad de 3,6Lm/h ¡que velocidad final tenia el perro si lla masa es de 8kg?
Physics
1 answer:
MatroZZZ [7]3 years ago
6 0

Responder:

6.704 m / s

Explicación:

Se dice que el trabajo se realiza cuando la fuerza aplicada a un objeto hace que el objeto se mueva. Primero necesitamos calcular la distancia recorrida por el perro usando la fórmula del trabajo realizado.

Trabajo realizado = Fuerza × distancia

Distancia = Trabajo realizado / Fuerza

Distancia = W / mg

S = 176/8 × 9,81

S = 176 / 78,48

S = 2,24 m

Dada la velocidad inicial u = 3.6km / h

Convertir a m / s

= 3.6km × 1000m / 1h × 3600

= 3600/3600

= 1 m / s

u = 1 m / s

Usando la ecuación de movimiento

v² = u² + 2gS para obtener la velocidad final v:

v² = 1² + 2 (9,81) (2,24)

v² = 1 + 43,9488

v² = 44,9488

v = √44,9488

v = 6,704 m / s

Por tanto, la rapidez final del perro es de 6,704 m / s

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⇒   ∆x = (20.0 m/s)² / (8.40 m/s²) ≈ 47.6 m

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Jennifer's cat is stuck in a
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The distance is 10m

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Required speed is 10/22=0.45 m/s

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2 years ago
Compare and contrast the strength of the forces between two objects with a mass of 1 kg each, a charge of 10
DochEvi [55]

Answer:

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

* The gravitational force is always attractive, the electrical force can be attractive or repulsive.

Explanation:

Let's start by calculating each force.

Gravitational force

             F =G \frac{m_1m_2}{r^2}  

let's calculate

             F = 6.67 10⁻¹¹  1  1 / 1²

             F = 6.67 10⁻¹¹ N

Electric force

             F = k \frac{q_1q_2}{r^2}  

indicates that the charge is q = 10 C

            F = 9 10⁹ 10 10 / 1²

            F = 9 10¹¹ N

Let's see the similarities between the two forces

* are proportional to the product of a magnitude, mass or charge

* They are inversely proportional to the square of the distance

* They are long-range forces since zero is not made up to an infinite distance. The gravitational force is always attractive, the electrical force can be attractive or repulsive.

The differences in them

* The electric force in much greater than the gravitational force

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A magnitic field is used to bend a beat of electrons. What uniform magnetic field is required to bend a beam of electrons moving
jarptica [38.1K]

The strength of the magnetic field is 2.73\cdot 10^{-5} T

Explanation:

When a charged particle is moving in a uniform magnetic field, the particle experiences a force perpendicular to the direction of motion. This force is given by

F=qvB

where

q is the charge of the particle

v is the velocity of the particle

B is the strength of the magnetic field

Since this force acts perpendicular to the direction of motion, the particle moves in a circular motion and the force acts as a centripetal force, so we can write:

F=qvB = m \frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the circular orbit

We can re-arrange the equation in order to isolate B:

B=\frac{mv}{qr}

In this problem, we have electrons, with

m=9.11\cdot 10^{-31} kg

v=1.2\cdot 10^6 m/s

q=1.6\cdot 10^{-19} C

r = 0.25 m

Substituting these numbers, we find the strength of the magnetic field:

B=\frac{(9.11\cdot 10^{-31})(1.2\cdot 10^6)}{(1.6\cdot 10^{-19})(0.25)}=2.73\cdot 10^{-5} T

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