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posledela
3 years ago
15

Two people are sitting on playground swings. One is pulled back 4 degrees from the vertical and the other is pulled back 8 degre

es. They are both released at the same instant. Will they both come back to their starting points at the same time
Physics
2 answers:
alex41 [277]3 years ago
5 0

Answer:

The bodies will not come back to their starting point at he same time.

Explanation:

Since they are both pulled back at an angle to the vertical, there is a tangential component of acceleration a = gsinθ

When θ = 4 , a = 9.8sin4 = 0.684 m/s²

When θ = 8 , a = 9.8sin8 = 1.364 m/s²

Using s = ut + 1/2at². Where s is the distance covered and t = time taken, u = initial speed = 0 (assumed since they are both released at the same time)

So s = 0 × t + 1/2at² = 1/2at²

s = 1/2at²

t = √2s/a. Now, since s is the same for both swings, it follows that

t ∝ 1/a. Since their accelerations are different, the bodies will not come back to their starting point at he same time.

lara [203]3 years ago
4 0

Answer:

They will come back at the same time.

Explanation:

The angular velocity equation of ω= \frac{V}{r} where ω is the frequency of the movement, dependent on the angle. But since swings are simple pendulums and their angles of 8 and 4 degrees are small, they will come back to their starting points at the same time.

I hope this answer helps.

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Who was the proponent of the Neo-classicism?

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2 years ago
Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
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Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) x = 4.47 cm

c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x = 1.48 cm

Explanation:

a) The center of mass is equal to:

x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

d₂ = 1.9 cm

d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm

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3 years ago
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Answer:

When you jump off a train, you jump off a certain height and your downwards (vertical) velocity is zero. But your forward (horizontal) velocity is not. You will hit the ground on split second with your horizontal velocity practically the same as the train.

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3 years ago
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What
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Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

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Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

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