Login

Search the forums now!

Welcome, Guest. Please login or register.

November 08, 2025, 06:55:39 am
Pages: [1]   Go Down

Author Topic: Help with questions  (Read 904 times)  Share 

0 Members and 1 Guest are viewing this topic.

Keyzer

  • Victorian
  • Trailblazer
  • *
  • Posts: 47
  • Respect: +1
Help with questions
« on: March 24, 2010, 05:03:23 pm »
0
A stationary billiard ball of mass 0.2 kg is hit by a cue, increasing its speed to 5 m/s. The total contact time between the cue and the ball is 0.002 seconds. Find:
Momentum
Impulse
Acceleration

A satellite of 7.34 x 10^22 kg orbits the earth. The radius between it and the earth is 3.0 x 10^8 m and the period of orbit is 2, 358, 720 seconds.
Find:
Orbital speed
Acceleration
Magnitude of force keeping satellite in motion
Logged

superflya

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1763
  • EL-Heat.
  • Respect: +8
Re: Help with questions
« Reply #1 on: March 24, 2010, 05:20:47 pm »
0
to calculate momentum p=mv.
the impulse = change in momentum
acceleration =

question2) simply plug in the values.



again just a matter of subbing in given values.
Logged
2010- English, Methods (CAS), Physics, Specialist, Chem.
2011- Bachelor of Commerce/Aerospace Engineering - Monash


"The day i stop learning, is the day i walk away from the game" Michael Jordan.

Keyzer

  • Victorian
  • Trailblazer
  • *
  • Posts: 47
  • Respect: +1
Re: Help with questions
« Reply #2 on: March 24, 2010, 05:58:31 pm »
0
Quote from: superflya on March 24, 2010, 05:20:47 pm
to calculate momentum p=mv.
the impulse = change in momentum
acceleration =

question2) simply plug in the values.



again just a matter of subbing in given values.

Could you use circular motion equations to figure out Question 2? For v, could you use v = 2piR/T ?
I mean, for acceleration, the formula is (v^2/ R), which also equals (4pi^2 R/T^2). I use this to solve for v, which gives me the same answer as v = 2piR/T.
« Last Edit: March 24, 2010, 06:13:44 pm by Keyzer »
Logged

qshyrn

  • Victorian
  • Forum Leader
  • ****
  • Posts: 535
  • Respect: +2
Re: Help with questions
« Reply #3 on: March 24, 2010, 06:50:20 pm »
0
yeah u can do it that way as well i think
Logged

Juddinator

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 269
  • Respect: -4
Re: Help with questions
« Reply #4 on: April 01, 2010, 11:24:03 am »
0
A mass of 25kg is on a rough horiz. floor. A horiz. force of 60N to the left acts on the mass. The total friction force opposing the motion is 5N.

What would be the acceleration if the force was applied at an angle of 45 to the horizontal?

Is the formula appropriate to use here?

Thanks
Logged

m@tty

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 4324
  • Respect: +33
  • School: Heatherton Christian College
  • School Grad Year: 2010
Re: Help with questions
« Reply #5 on: April 01, 2010, 11:36:06 am »
0
I don't know what formula that is....

The new horizontal component of the 60N force at an angle of is

So in the leftward direction

And

So,

to the left

EDIT: UNITS!!
« Last Edit: April 01, 2010, 12:28:41 pm by m@tty »
Logged
2009/2010: Mathematical Methods(non-CAS) ; Business Management | English ; Literature - Physics ; Chemistry - Specialist Mathematics ; MUEP Maths

96.85

2011-2015: Bachelor of Aerospace Engineering and Bachelor of Science, Monash University

2015-____: To infinity and beyond.

Juddinator

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 269
  • Respect: -4
Re: Help with questions
« Reply #6 on: April 01, 2010, 12:24:14 pm »
0
Quote from: m@tty on April 01, 2010, 11:36:06 am
I don't know what formula that is....

The new horizontal component of the 60N force at an angle of is

So

And

So,


Sweet, thanks for that m@tty!
Logged
Pages: [1]   Go Up