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Lapatulllka [165]
3 years ago
5

How much work did she perform?

Physics
1 answer:
Vitek1552 [10]3 years ago
7 0
She performed a lot of work
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Professional Application. A 96 kg football player catches a 0.900 kg ball with his feet off the ground with both of them moving
Zarrin [17]

To solve this problem it is necessary to apply the equations related to the conservation of momentum.

This definition can be expressed as

m_1u_1+m_2u_2 = (m_1+m_2)V_f

Where

m_{1,2} = Mass of each object

u_{1,2} = Initial Velocity of each object

V_f= Final velocity

Rearranging the equation to find the final velocity we have,

V_f = \frac{m_1u_1+m_2u_2}{(m_1+m_2)}

Our values are given as

m_1 = 96Kg\\m_2 = 0.9Kg\\u_1 = 6.3m/s\\u_2 = 27.4m/s

Replacing we have,

V_f = \frac{(96)(6.3)+(0.9)(27.4)}{(96+0.9)}

V_f = 6.4959m/s

Therefore the final velocity is 6.5m/s

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3 years ago
Are individual organelles (cell structures) living? Explain
Julli [10]
All cells are living however organelles are not living there are no organisms that consist of just a single cell
3 0
4 years ago
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Please help, I do not understand
Anettt [7]
I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed =  ω  rad/sec
Time per unit angle =  (1/ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed =  2ω  rad/sec
Time per unit angle =  (1/2ω)  sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good.  We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed.  Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

     (a fraction of one orbit) = (the same fraction of the other orbit)    
AND
     the total elapsed time is a common multiple of their periods.

Wait !  Ignore all of that.  I'm doing a good job of confusing myself, and
probably you too.  It may be simpler than that.  (I hope so.)  Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.  
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed.  We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds)  =  (K+1) (π/ω seconds)

                     2Kπ/ω   =    Kπ/ω + π/ω

Subtract  Kπ/ω :    Kπ/ω = π/ω

Multiply by  ω/π :      K  =  1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

     In the time it takes the slower planet to revolve once,
     the faster planet revolves twice, and catches up with it.
    
     It will be  2π/ω  seconds before the planets line up again.
    
     When they do, they are again in the same position as shown
     in the drawing.

To describe it another way . . . 

     When Kanye has completed its first revolution ...

     Bieber has made it halfway around.

     Bieber is crawling the rest of the way to the starting point while ...

     Kanye is doing another complete revolution.

     Kanye laps Bieber just as they both reach the starting point ...

     Bieber for the first time, Kanye for the second time.


You're welcome.  The generous bounty of 5 points is very gracious,
and is appreciated.  The warm cloudy water and green breadcrust
are also delicious.
5 0
3 years ago
Which of the following is an example of Newton's Third Law?* O A stack of pennies will not move unless you flick them over. O Fa
Morgarella [4.7K]

Answer:

A ball hits the ground and the ground pushes up on it

Explanation:

Newton's third law basically states that for every action, there's a reaction.

a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.

Hope this Helps!!! :)

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Which statement is true of cooling down after physical activity? A cool-down is a period of semistrenuous activity after physica
Xelga [282]
The statement that is true of cooling down after physical activity is that you should cool down for about 5 to 10 minutes after being physically active.
8 0
4 years ago
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