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MrBill31
MrBill31, Engineer
Category: Homework
Satisfied Customers: 85
Experience:  BS General Physics and Chemistry MS Physics specialty biophysics
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A hockey puck with mass 0.160kg is at rest on the ...

Customer Question

A hockey puck with mass 0.160kg is at rest on the horizontal, frictionless surface of a rink. A player applies a force of 0.250N to the puck, parallel to the surface of the ice, and continues to apply this force for 2.00s .What is the position of the puck at the end of that time?
Submitted: 6 years ago.
Category: Homework
Expert:  MrBill31 replied 6 years ago.

F = ma

0.25N = 0.160kg * a

solve for a

a = 0.25(kg m/s2)/0.160 kg = 1,5625 m/s2

If the puck is at rest and then accelerates at a constant rate the position is:

x = (1/2)at2 = (1/2)1,5625 m/s2 * (2 s)2

x = 3.125 meters away from the start position

Hope this helps

-MrBill

MrBill31, Engineer
Category: Homework
Satisfied Customers: 85
Experience: BS General Physics and Chemistry MS Physics specialty biophysics
MrBill31 and 4 other Homework Specialists are ready to help you
Customer: replied 6 years ago.
i did that problem already i wanted to know how to find the speed of the puck at the end of time.
Expert:  MrBill31 replied 6 years ago.

the velocity is given by

v = at

so since a = 1,5625 m/s2

we have

v = at = 1,5625 m/s2 * 2s = 3.125 m/s

WARNING the fact that these two problems have the same numerical value is only a cooincidence!

Customer: replied 6 years ago.
Thank you is it $15 for every ? we ask?
Expert:  MrBill31 replied 6 years ago.

I guess there is the bid charge (at least on your end) for every question you ask.

Sorry I don't make the rules.

I guess that means you have to be careful what you ask.

I have seen people post several problems as part of a single question. The risk of doing that is if the assignment is too long you may not get a response.

Still $15 for that 2 line answer IS a bit steep.

Next time I'd post both questions together as a single question.

Again my apologies

Customer: replied 6 years ago.
When venturing forth on Planet X, you throw a 5.24kg rock upward at 13.0m/s and find that it returns to the same level 1.51s later. what does the rock weigh on planetX?
Expert:  MrBill31 replied 6 years ago.

throw rock up into air

we know

v = at + vo

where a is the acceleration and vo is the initial velocity

I like to look at this like this. The rock spends 1/2 it's time going up and 1/2 it's time returning to the same level

so the time going up is 1.51s/2 = 0.755 s

The rock decellerates from 13.0 m/s to 0 m/s in this amount of time so

a = dv/dt = (13.0m/s)/(0.755 s2) = 17.22 m/s2

(about 1.75 x earths gravity)

What does the rock weigh ?

in Newtons it would be F = mgplanet x = 5.24kg*17.22 m/s2

= 90.2 N

hope this helps

- MrBill

 

Customer: replied 6 years ago.
A tennis ball traveling horizontally at 22.0m/s suddenly hits a vertical brick wall and bounces back with a horizontal velocity of 18.0m/s.make a free body diagram before it hits the wall, just after it has bounced free of the wall and while in contact witht the wall
Expert:  MrBill31 replied 6 years ago.
Note: drawings take longer
Customer: replied 6 years ago.
okay
Expert:  MrBill31 replied 6 years ago.
THIS ANSWER IS LOCKED!

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