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Inessa [10]
2 years ago
14

A golf ball is launched horizontally at a speed of 11 meters per second and a high of 6.4 m above the ground. How long will it t

ake for the golf ball to reach the ground?

Physics
1 answer:
Serjik [45]2 years ago
8 0

The answer is A.

t = ( 2h / g )^1/2 = ( 2 x 6.4 / 9.8 )^1/2 = 1.14s

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If there is 4% increase in orbital radius...then by how much its time period will increase?
erastovalidia [21]

Kepler's 3rd law of planetary motion is exactly
what we need in order to answer this one:

      (orbital period)² / (orbital radius)³ =
       the same number for all bodies orbiting the sun

Let's call that number 'K' just for convenience.

So we know that  T₀² / R₀³  =  K

We're going to be looking for 'T', so let's rearrange the equation now.
Multiply each side by  R₀³ .
Now it says
                             T₀²  =  K R₀³

Now, take the square root of each side, and we have
 
                             T₀  =   √ (K R₀³)  .

Now the radius is increased to  (1.04 R₀).
We want to find the new T .  

                            T  =   √ K · (1.04 R₀)³

                                =    √ K · 1.124864 R₀³

Pull that decimal out of the radical, by taking its square root:

                               =  1.0606 √K · R₀³       

                           T  =   1.0606  T₀

The orbital time has increased by 6%  .
_________________________________________

I suspect we probably could have said that since T  varies
as  R^1.5 power, we should look for  (1.04)^1.5 power.

             (1.04)^1.5  =  1.0606  <== bada-bing




3 0
3 years ago
A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms2.75 ms from an init
kumpel [21]

Answer:

2677.3 N

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v  = final speed of the hand = 0 m/s

m = Total mass of hand and forearm = 1.55 kg

t = time interval for hand to come to rest = 2.75 ms = 0.00275 s

F = Force applied on the leg

Using Impulse-change in momentum equation

F t = m (v - v₀)

F (0.00275) = (1.55) (0 - 4.75)

F = - 2677.3 N

magnitude of force = 2677.3 N

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