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drek231 [11]
3 years ago
6

Find the magnitude of the force needed to accelerate a 300 g mass with a⃗ = -0.205 m/s2 i^+0.700 m/s2 j

Physics
1 answer:
never [62]3 years ago
7 0
F=ma
m=300g= 0.3kg
a= -0.205i +0.700j m/s2 = <-0205, 0.7>m/s2

Thus, F = 0.3<-0.205, 0.7> N
F = <-0.0615, 0.21> N
= -0.0615i N +0.2100j N
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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.
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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

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we also know that

s=76 cm= 0.76 m ( maximum s)

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

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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

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Explanation:

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