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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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Answer:

The magnitude of F when t=4 s=146.64 N

Explanation:

We are given that

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Height of crate above its initial position is given by

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We have to find the magnitude of F when t=4.00 s

Differentiate w.r.t t

\frac{dy}{dt]=2.8+3(0.61)t^2

\frac{d^2y}{dt^2}=6(0.61)t

a(t)=\frac{d^2y}{dt^2}=6(0.61)t m/s

a(4)=6(0.61)(4)=14.64 m/s^2

Now, magnitude of force

F=m(a+g)=6(14.64+9.8)=146.64N

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3 years ago
A net force of 50 N causes a mass to accelerate at a rate of 6.8 m/s2. Determine the mass. ​
Taya2010 [7]
  • Force=50N
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\\ \sf\longmapsto F=ma

\\ \sf\longmapsto m=\dfrac{F}{a}

\\ \sf\longmapsto m=\dfrac{50}{6.8}

\\ \sf\longmapsto m=7.3kg

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Explanation:

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You use the following formula for the range of the trajectory:

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You do v_oy the subject of the formula ( 1 ) and you replace the values of the other parameters in order to calculate v_oy:

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hence, the initial vertical velocity of the tiger is 16.33m/s

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