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yan [13]
3 years ago
14

1.Una partícula de masa m = 0.2 [kg] está unido a dos resortes idénticos (de longitud L = 1.2 [m]) sobre la parte superior de un

a mesa horizontal sin fricción. Los dos resortes tienen la misma constante elástica de K = 40 [N/m] y cada uno esta inicialmente en su posición de equilibrio (punto O). Si la partícula se hala una distancia d = 0.5 [m] hacia la derecha y después se suelta (ver figura), ¿cuál es la velocidad cuando pasa por el punto O (posición de equilibrio)?
Physics
1 answer:
uysha [10]3 years ago
8 0

Answer:

walang kuwento

Explanation:

kasi ma answer ko

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What is the acceleration of a 10kg mass pushed by a 5N force?​
GenaCL600 [577]

Answer:

<h2>0.5 m/s²</h2>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

From the question we have

a =  \frac{5}{10} =  \frac{1}{2}   \\  = 0.5

We have the final answer as

<h3>0.5 m/s²</h3>

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5 0
3 years ago
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
loris [4]

Answer: 36.86\°

Explanation:

According to the described situation we have the following data:

Horizontal distance between lily pads: d=2.4 m

Ferdinand's initial velocity: V_{o}=5 m/s

Time it takes a jump: t=0.6 s

We need to find the angle \theta at which Ferdinand jumps.

In order to do this, we first have to find the <u>horizontal component (or x-component)</u> of this initial velocity. Since we are dealing with parabolic movement, where velocity has x-component and y-component, and in this case we will choose the x-component to find the angle:

V_{x}=\frac{d}{t} (1)

V_{x}=\frac{2.4 m}{0.6 s} (2)

V_{x}=4 m/s (3)

On the other hand, the x-component of the velocity is expressed as:

V_{x}=V_{o}cos\theta (4)

Substituting (3) in (4):

4 m/s=5 m/s cos\theta (5)

Clearing \theta:

\theta=cos^{-1} (\frac{4 m/s}{5 m/s})

\theta=36.86\° This is the angle at which Ferdinand the frog jumps between lily pads

4 0
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Likurg_2 [28]

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