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i got sum online questions to do.... but i have no idea what to do. maybe sum of u physics whizzes can help me out
(i.e. point me in the right direction)
1.A battery has an emf of 15.0 V. The terminal voltage of the battery is 13.6 V when it is delivering 20.0 W of power to an external load resistor R.
(a) What is the value of R?
(b) What is the internal resistance of the battery?
2.
http://www.webassign.net/sj/p21-31.gif
Three 100 resistors are connected, as shown in Figure P21.30. The maximum power that can safely be dissipated in any one resistor is 26.5 W.
(a) What is the maximum voltage that can be applied to the terminals a and b?
(b) For the voltage determined in part (a), what is the power dissipated in each resistor?
Resistor on the left
Resistor at the top of the loop
Resistor at the bottom of the loop
What is the total power dissipation?
3. Consider the circuit shown in Figure P21.27 (R = 32.0)
http://www.webassign.net/sj/p21-29alt.gif
(a) Find the current in the 32.0 resistor.
(b) Find the potential difference between points a and b.
Thanks
50
_________________
Wow 1800+ posts....
[url=http://www.traders-guild.co.uk" TARGET="_blank]>>*Trader's Guild*[ This Message was edited by: 50Cent on 2005-02-13 15:34 ]
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Posted: 2005-02-13 16:30:45
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Seems like a basic electronics class.. Read the book and answer the questions.. Im sure your teacher wouldn't want you to be asking on the net for the answers to his/her questions.
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Posted: 2005-02-13 17:43:32
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Here you go:
NOTE: Its almost mornin here n i'm tired so some errors may have inadvertently creeped in to i doubt it
Ans 1)
(a) 9.24
(b) 0.95
Ans 2)
(a) 77.21V
(b) 26.5V
6.62V
6.62V
39.75V
Ans 3)
(a) 0.15A
(b) 5.837V
Hope i've solved them correctly
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Posted: 2005-02-14 02:20:54
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reminds me of high schools physics...
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Posted: 2005-02-14 08:00:51
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Reminds me how dumb at science am i
This message was posted from a T610
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Posted: 2005-02-14 08:08:33
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You should be able to figure it out, they're just basic Resistor circuits. You'll need the following furmalae:
P=(I^2)R=(E^2)/R; E=IR
Series Resistors: Rt=R1+R2+R3...+Rn
Parallel Resistors: Rt=(R1xR2)/(R1+R2)
edit: if R is more than 2 in parallel:
Rt=1/((1/R1)+(1/R2)...+(1/Rn))
Tip: Simplify your circuit first like in No. 2, Solve the parallel first, then the series.
_________________
resistance is futile.
[ This Message was edited by: blackspot on 2005-02-14 10:52 ]
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Posted: 2005-02-14 08:29:42
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nice one guys!!
100%
Thanks
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Posted: 2005-02-14 10:33:54
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1. A block of mass m = 2.20 kg slides down a 30.0° incline which is 3.60 m high. At the bottom, it strikes a block of mass M = 7.30 kg which is at rest on a horizontal surface, Fig. 7-41. (Assume a smooth transition at the bottom of the incline, an elastic collision, and ignore friction.)
http://www.webassign.net/giancoli5/7_41.gif
(a) Determine the speeds of the two blocks after the collision.
lighter block
(b)heavier block
(c) Determine how far back up the incline the smaller mass will go.
2. A 1.0 103 kg Rover collides into the rear end of a 2.2 103 kg BMW stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.30, calculates the speed of the Rover at impact. What was that speed?
3. An 18 g rifle bullet traveling 215 m/s buries itself in a 3.7 kg pendulum hanging on a 2.8 m long string, which makes the pendulum swing upward in an arc. Determine the horizontal component of the pendulum's displacement.
A 0.300 kg croquet ball makes an elastic head-on collision with a second ball initially at rest. The second ball moves off with half the original speed of the first ball.
(a) What is the mass of the second ball?
(b) What fraction of the original kinetic energy (KE/KE) gets transferred to the second ball?
I dun most of them... but these are just beyond me :S
i dont even think they're in my syllabus
Thanks
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Posted: 2005-03-07 18:30:16
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Oke here you go again
Any careless errors that may have creeped in regretted
Ans 1)
(a) 4.50m/s
(b) 3.90m/s
(c) it attains a height of 1.03m which is equivalent to moving back along the surface by a distance of 2.06m
Ans 2)
I cant figure out the values of the masses you've written so i'm taking the rover's mass as "m" and that of the BMW as "M" You just substitute the respective values in this equation to get the value of "v"...
(0.30) x (m+M) x 9.8 = 0.5 x m x (v^2)
Ans 3) 1.11m
Ans 4)
(a) 0.9kg
(b) 0.75 or 3/4
Adios!!
[ This Message was edited by: clank on 2005-03-07 20:13 ]
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Posted: 2005-03-07 21:10:00
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Whey Clank! Well done mate!
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Posted: 2005-03-07 21:16:23
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