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Whitepunk [10]
3 years ago
13

Una persona y su perro se encuentran inicialmente en reposo y paradas arriba de una caja que está apoyada sobre un piso rugoso.

La masa total de la caja, la persona y la mascota es de 80 kg. El coeficiente de rozamiento cinético entre la caja y el piso vale 0,3. Cuando el perrito salta hacia el piso con una velocidad horizontal de 6,0 m/s, la caja y la persona se desplazan 1,0 m en sentido contrario. ¿Cuál es la masa de la mascota?
Physics
1 answer:
alexandr1967 [171]3 years ago
5 0

Answer:

La masa de la mascota es de 11,24 kg.

Explanation:

Dado que la masa total de = 80 kg

Tenemos

Fuerza de fricción = 80 kg × 9.81 × 0.3 = 235.44 N

La mascota salta horizontalmente con una velocidad, v, de 6.0 m / s para dar un movimiento de 1 m a la caja de masa y a la persona

Trabajo realizado por el salto de la mascota = (80 - m) × 9.81 × 0.3

La energía del movimiento del perro = 1/2 × m × v² = Trabajo realizado al saltar de la mascota

1/2 × m × 6² = (80 - m) × 9.81 × 0.3

20943 · m = 235440

m = 235440/20943 = 11,24 kg

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A string under a tension of 36 N is used to whirl a rock in a horizontal circle of radius 3.6 m at a speed of 16.12 m/s. The str
ExtremeBDS [4]

Answer:

6010.457N

Explanation:

Centripetal acceleration = a= V²/R

At a radius of 3.6m and velocity of 16.12m/s,

Acceleration is

a = 16.12²/ 3.6 = 72.182 m/s²

Force = Mass (m) * Acceleration (a)

36 = m * 72.182

m = 36/72.182

At breaking point

Radius = 0.468 m and Velocity = 75.1 m/s

a = V²/R = 75.1²/0.468

a = 12051.3 m/s

F = Mass(m) * Acceleration (a)

F = m * 12051.3

m = F/ 12051.3

Settings the ratio of mass equal

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=> 36/72.182 = F/12051.3

F = 12051.3 * 36/72.182

F = 6010.457N

3 0
3 years ago
An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2. The coil rotates in a uniform magnetic f
Reptile [31]

Answer:

331.75 V

Explanation:

Given:

Number of turns of the coil, N = 40 turns

Area, A = 0.06 m²

Magnetic Field, B = 0.4 T

Frequency, f = 55 Hz

                           Maximum induce emf, E₀ = NABω

but ω = 2πf

                           Maximum induce emf, E₀ = NAB(2πf₀)

                           Maximum induce emf, E₀ = 2πNABf₀

Where;

N is number of turns of the coil

A is area

B is magnetic field

ω is the angular velocity

f is the frequency

                                     E₀ = 2 × π × 40 × 0.06 × 0.4 × 55

                                     E₀ = 342.81 V

The maximum induced emf is 331.75 V

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