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slega [8]
3 years ago
5

Alicia está a punto de perder su bus. En un desesperado intento, corre a una velocidad constante de 5 m/s. Cuando está a 15 m de

la parada, el bus arranca con una aceleración constante de 0,5m/s 2 . ¿Logrará Alicia alcanzar el bus?
Physics
1 answer:
pochemuha3 years ago
7 0

Answer:

Si logra alcanzar el bus.

Explanation:

Para poder solucionar este problema debemos de tener en cuenta que Alicia corre a velocidad constante para poder alcanzar el bus. La formula de la cinematica que tiene en cuenta la velocidad constante es la siguiente:

x_{f} = x_{o}+(v*t)

donde:

Xf = Ubicacion del punto donde se encuentra el bus [m]

Xo = Ubicacion desde donde esta Alicia [m]

v = velocidad constante = 5 [m/s]

t = tiempo [s]

Xf - Xo = 15 [m]

15 = 5*t

t = 3 [s]

Ahora con el tiempo podemos encontrar la velocidad del bus por medio de la siguiente ecuacion de cinematica para la aceleracion constante:

v_{f} = v_{i}+(a*t)

donde:

Vf = velocidad del bus despues de los 3 [s]

Vi = velocidad inicial = 0

a = aceleracion = 0.5 [m/s^2]

Vf = 0 + (0.5*3)

Vf = 1.5 [m/s]

La velocidad del bus es menor que la velocidad de Alicia, por ende Alicia alcanzara el bus.

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5 0
3 years ago
Two particles collide, one of which was initially moving and the other initially at rest. Is it possible for both particles to b
solmaris [256]

Answer:

Not possible

Explanation:

Unless there's some extra external force to keep both particles at rest after the collision, the momentum must be conserved before and after the collision.

So before the collision, 1 particle is at rest, 1 not ->  total momentum is non-zero

After the collision, both particles are at rest -> total momentum is zero which is different from before.

Therefore this is not possible.

4 0
3 years ago
A 2.61 g lead weight, initially at 11.1 ∘C, is submerged in 7.67 g of water at 52.6 ∘C in an insulated container. What is the fi
sleet_krkn [62]

Answer:

Equilibrium temperature will be T=52.2684^{\circ}C

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So 2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)

0.334T-3.67=1688-32.031T

T=52.2684^{\circ}C

5 0
2 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

  = 207 × 1.342 + 1/2 × 896.65 × 1.342²

  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

5 0
3 years ago
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