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3 years ago

6

A stone is thrown vertically upward with a speed of 18.0 m/s. How fast is it moving when it reaches a height of 11.0m? How long

does it take to reach the height?

1 answer:

steposvetlana [31]3 years ago

8 0

Given: v0= 18.0 m/s, y0=0m, yf=11m, g=-9.81 m/s^2

v0= initial velocity, vf= final velocity, y0= initial height, yf= final height, g= gravity, sqrt()= square root, ^2=squared

vf^2=v0^2 + (2)(g)(yf-y0)
vf^2=(18.0 m/s)^2+(2)(-9.81 m/s^2)(11 m-0m)
vf^2=18.0 m/s)^2 + (-19.62 m/s^2)(11 m)
vf^2=(324 m^2/s^2) - (215.82 m^2/s^2)
vf^2=108.18 m^2/s^2
vf=sqrt(108.18 m^2/s^2)
vf=10.4 m/s

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Answer:

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$\frac{d{Fe} }{ d{Cu} } =\sqrt{\frac{p_{Fe} }{ p_{Cu} }}$

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from the question the two materials have the same resistance per unit length.

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re-arranging the equation we have,

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A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

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Answer :

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where,

v = final velocity = 5 m/s

u = initial velocity = 25 m/s

t = time = 10 s

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Now put all the given values in the above equation 1, we get:

$5m/s=25m/s+a\times (10s)$

$a=-2m/s^2$

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By the 2nd equation of motion,

$s=ut+\frac{1}{2}at^2$ ...........(2)

where,

s = distance covered by the car = ?

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a = acceleration of the car = $-2m/s^2$

Now put all the given values in the above equation 2, we get:

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