Webassign 4
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Purple Squirrel and Blue Angel
READ FIRST: THIS WEBSITE (www.myspace.com/webassign or www.webassign.tk) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. LEGEND IN EQUATIONS (WORK SHOWN): BLUE—COMMON NUMBERS (same for everyone) GREEN—FOUND IN EXPLANATION PURPLE—YOUR PERSONAL NUMBERS RED—ANSWER
1. A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8400 N/C. The mass of the water drop is 3.80 10-9 kg. a) The water drop is suspended in the air due to the opposing force of the electric
field below it. The electric field has a POSITIVE charge of 8480N/C; therefore the charge of the drop of water must also be positive. (*Note: Water does not have charges, but in this case the water must have ions of some sort.) b) We (Purple Squirrel and Blue Angel) are currently working on this part of the
problem, check back for the updated version of this explanation at 6:00 P.M. We are sorry for any inconveniences.
2. A small object, which has a charge q = 6.9 µC and mass m = 8.7 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 103 m/s in a time of 0.84 s. Determine the magnitude of the electric field.
Σ F = ma Σ F = Eq As you see above, the total force is equal to the mass times the acceleration, the force is also equal to the Electric Field Strength times the charge. If you set them equal to each other you get: ma = Eq (ma) / q = E Use the both the speed and the time that are given to find the acceleration. (2.0 x 103) / (.84) = 2380.95 Use the acceleration that you just found along with the numbers given in the problem to plug them in and solve. ((8.7 x 10-5) x (2380.95)) / (6.9 x 10-6) = 30020.70N/C This WebAssign was brought to you by:
Blue Angel
THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS.
READ FIRST: THIS WEBSITE (www.myspace.com/webassign or www.webassign.tk) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS. LEGEND IN EQUATIONS (WORK SHOWN): BLUE—COMMON NUMBERS (same for everyone) GREEN—FOUND IN EXPLANATION PURPLE—YOUR PERSONAL NUMBERS RED—ANSWER
1. A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8400 N/C. The mass of the water drop is 3.80 10-9 kg. a) The water drop is suspended in the air due to the opposing force of the electric
field below it. The electric field has a POSITIVE charge of 8480N/C; therefore the charge of the drop of water must also be positive. (*Note: Water does not have charges, but in this case the water must have ions of some sort.) b) We (Purple Squirrel and Blue Angel) are currently working on this part of the
problem, check back for the updated version of this explanation at 6:00 P.M. We are sorry for any inconveniences.
2. A small object, which has a charge q = 6.9 µC and mass m = 8.7 10-5 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 103 m/s in a time of 0.84 s. Determine the magnitude of the electric field.
Σ F = ma Σ F = Eq As you see above, the total force is equal to the mass times the acceleration, the force is also equal to the Electric Field Strength times the charge. If you set them equal to each other you get: ma = Eq (ma) / q = E Use the both the speed and the time that are given to find the acceleration. (2.0 x 103) / (.84) = 2380.95 Use the acceleration that you just found along with the numbers given in the problem to plug them in and solve. ((8.7 x 10-5) x (2380.95)) / (6.9 x 10-6) = 30020.70N/C This WebAssign was brought to you by:
Blue Angel
THIS WEBSITE (www.myspace.com/webassign) WILL BE OPEN TO THE PUBLIC ONLY FOR THE FIRST TWO WEBASSIGNS. CHANGES TO THE NUMBER OF WEBASSIGNS BEFORE THE WEBSITE CLOSES, IS SUBJECT TO CHANGE AT OUR DISCRETION. THIS WEBSITE WILL BE CLOSED TO PREVENT TEACHERS COMING IN. IF YOU WANT TO HAVE ACCESS TO THE WEBSITE AFTER IT “CLOSES” YOU MUST REQUEST US TO ADD YOU AS FRIENDS.
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