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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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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend.
MatroZZZ [7]

Answer: Part(a)=0.041 secs, Part(b)=0.041 secs

Explanation: Firstly we assume that only the gravitational acceleration is acting on the basket ball player i.e. there is no air friction

now we know that

a=-9.81 m/s^2  ( negative because it is pulling the player downwards)

we also know that

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using kinetic equation

v^2=u^2+2as

where v is final velocity which is zero at max height and u is it initial

hence

u^2=-2(-9.81)*0.76

u=3.8615 m/s\\

now we can find time in the 15 cm ascent

s=ut+0.5at^2

0.15=3.861*t+0.5*9.81t^2\\

using quadratic formula

t=\frac{-3.861+\sqrt{3.86^2-4*0.5*9.81(-0.15)} }{2*0.5*9.81}

t=0.0409 sec

the answer for the part b will be the same

To find the answer for the part b we can find the velocity at 15 cm height similarly using

v^2=u^2+2as

where s=0.76-0.15

as the player has traveled the above distance to reach 15cm to the bottom

v^2=0^2 +2*(9.81)*(0.76-0.15)

v=3.4595

when the player reaches the bottom it has the same velocity with which it started which is 3.861

hence the time required to reach the bottom 15cm is

t=\frac{3.861-3.4595}{9.81}

t=0.0409

8 0
3 years ago
How are gravity and electromagnetic force similar and different?
oksano4ka [1.4K]

Answer:

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Explanation:

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Explanation:

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