PHYS 110 - Physics Test 3
PHYS 110 - Physics Test 3
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PHYS 110
Grossmont College
FALL, 2015
Campus: Grossmont
Department of Physics
Test 3
(Time allowed: DUE 11:35 AM, MONDAY 12/14/2015)
NOTE:
1. Attempt all questions.
2. You may get help (not answers) from tutors and/or classmates.
3. Make sure you show your work, or give an explanation of how you came to your
answer. If there is no work or explanation there will be NO POINTS.
1. A force with magnitude F is needed to separate two charges by a distance r. What will be
1
the new force F ′ , needed to keep the two charges separated by a distance 2 r?
2. A a particle with charge of Q = 3.2 × 10−6 C is stuck to the top of a table. If another particle
with charge of q = 7.7 × 10−7C, and mass m = 0.10kg is floating on top of the particle with
charge Q; how high above the table is the particle with charge q?
3. You want to make a particle accelerator in your house. You are going to use a power source
to charge two parallel plates as shown in Figure 1. The particles will then be accelerated
from one plate to the other. The particles you want to accelerate have a mass of m1 =
3.5 × 10−12 kg, and a charge of q = 8.5 × 10−6C. You would like to accelerate the particles
to a speed of v = 6.0 × 106 m .
s
(a) What should be the potential difference across the two parallel plates?
(b) Now you want to accelerate different particles (with out changing the potential difference
across the plates). The new particles have a mass of m2 = 1.4 × 10−13 kg, and the same
charge as the first particles. If the distance between the plates is d = 0.15m, what is
the force needed to accelerate one of the new particles?
(c) What will be the speed of the new particles when they reach the other side?
CONTINUED
–2–
PHYS 110
4. Figure 2 shows a circuit with a resistor R = 1.0 × 103 Ω, and a battery with potential
difference of VB = 5.0V .
(a) Solve the circuit in Figure 2. That is, find all unknown currents, voltages, and resistances.
(b) What is the direction of the current? Note: I want the direction of the conventional
current.
5. Consider the circuit in Figure 3, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) What is the equivalent resistance Req of the circuit?
(b) Solve the circuit. That is find all unknown currents, voltages, and resistances.
6. Consider the circuit in Figure 4, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) Find the equivalent resistance Req of the circuit.
(b) Draw the simplified equivalent circuit.
(c) Solve the circuit. That is, find all unknown currents, resistances, and voltages.
7. You want to build a circuit which causes a ligh tbulb to turn on when you throw a switch.
So, you build the circuit if Figure 5. When the switch is connected to point A the capacitor
charges, and when you connect the switch to point B the light goes on. Let the resistance of
the light bulb be Rℓ = 1.50 ×103 Ω, the potential difference across the battery is VB = 10.0V ,
and the capacitor has a capacitance of C = 1.35 × 10−4 F .
(a) What should be the resistance of R1 , if we want the capacitor to charge to 90% of its
maximum charge in a time t = 0.500s
(b) How long from the time you start to discharge the capacitor, does it take for the current
through the light bulb to get to 33.3% of its initial current?
8. Scientist use a device called a mass spectrometer to identify charged sub-atomic particles.
They send the particles with a known velocity, into a device containing a known uniform
magnetic field, causing the particle to move in a circle as shown in Figure 6. By
Physics
Name
Institution
Answer question 1
Force between charges is computed through the use of Coulombs law which indicates that F=kqQ/r2. The constant k = 8.99*109Nm2/C2. Force is normally a vector . when more than one charge exerts force on another charge, the sum vector of such charge is the vector summation of the individual forces. Such charge is similar to gravitational forces between the interacting masses. In both cases, the force is computed as 1/r2. This implies that a force used to separate charges by 1/2r would be computed as (1/2r)2.
Answer to question 2
When a particle with charge Q=3.2*10-6c is stuck on a table and another with charge q= 7.7*10-7C and with mass m= 0.10 kg floating on the same article bearing charge Q, the distance of the article with charge q could be computed. The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2
R= 555.78
Answer to question 3
• Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s
• Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons
The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1
Answer to question 4
Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps
B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below
Answer to question 5
• This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2. = 306.5.
• With voltage of 10, the currrent will be equal to voltage/resistance = 10/306.5= 0.03
Answer to question 6
• equivalent resistance of the ciruit = R1+R2/2 = 5.0*102+1.0*103=750
• The circuit is as indicated below
• The current = voltage/resistance = 10/750= 0.013
Answer to question in 7
•
E= V^2*C/2
Input:
Voltage across capacitor (V)
Capacitance (uF)
Load Resistance (Optional) (Ohms)
b.
Energy (Joules)
Time Constant (seconds)
Source: International Conference on Mechatronics Engineering and Electrical Engineering, & Sheng, (2015).
Answerr to question 8
• The detector should beplaced on the left side as the particle would be mopving in rounds towards the left side
• If the charge to mass ratio of the particle is q/m = 9.58*107C/KG and the velocity is 3.25*106,, the radius of the protons path would be caculated through the use of the formula as 9.58*107C/3.25*106 = 2.94769^13
International Conference on Mechatronics Engineering and Electrical Engineering, & In Sheng, A. (2015). Mechatronics engineering and electrical engineering: Proceedings of the 2014 International Conference on Mechatronics Engineering and Electrical Engineering (CMEEE 2014), Sanya, Hainan, P.R. China, 17-19 October 2014.
Comments
Content
PHYS 110 - Physics Test 3
Click Link Below To Buy:
http://hwcampus.com/shop/phys-110-physics-test-3/
Or Visit www.hwcampus.com
PHYS 110
Grossmont College
FALL, 2015
Campus: Grossmont
Department of Physics
Test 3
(Time allowed: DUE 11:35 AM, MONDAY 12/14/2015)
NOTE:
1. Attempt all questions.
2. You may get help (not answers) from tutors and/or classmates.
3. Make sure you show your work, or give an explanation of how you came to your
answer. If there is no work or explanation there will be NO POINTS.
1. A force with magnitude F is needed to separate two charges by a distance r. What will be
1
the new force F ′ , needed to keep the two charges separated by a distance 2 r?
2. A a particle with charge of Q = 3.2 × 10−6 C is stuck to the top of a table. If another particle
with charge of q = 7.7 × 10−7C, and mass m = 0.10kg is floating on top of the particle with
charge Q; how high above the table is the particle with charge q?
3. You want to make a particle accelerator in your house. You are going to use a power source
to charge two parallel plates as shown in Figure 1. The particles will then be accelerated
from one plate to the other. The particles you want to accelerate have a mass of m1 =
3.5 × 10−12 kg, and a charge of q = 8.5 × 10−6C. You would like to accelerate the particles
to a speed of v = 6.0 × 106 m .
s
(a) What should be the potential difference across the two parallel plates?
(b) Now you want to accelerate different particles (with out changing the potential difference
across the plates). The new particles have a mass of m2 = 1.4 × 10−13 kg, and the same
charge as the first particles. If the distance between the plates is d = 0.15m, what is
the force needed to accelerate one of the new particles?
(c) What will be the speed of the new particles when they reach the other side?
CONTINUED
–2–
PHYS 110
4. Figure 2 shows a circuit with a resistor R = 1.0 × 103 Ω, and a battery with potential
difference of VB = 5.0V .
(a) Solve the circuit in Figure 2. That is, find all unknown currents, voltages, and resistances.
(b) What is the direction of the current? Note: I want the direction of the conventional
current.
5. Consider the circuit in Figure 3, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) What is the equivalent resistance Req of the circuit?
(b) Solve the circuit. That is find all unknown currents, voltages, and resistances.
6. Consider the circuit in Figure 4, where R1 = 5.00×102Ω, R2 = 1.00×103Ω, and VB = 10.0V .
(a) Find the equivalent resistance Req of the circuit.
(b) Draw the simplified equivalent circuit.
(c) Solve the circuit. That is, find all unknown currents, resistances, and voltages.
7. You want to build a circuit which causes a ligh tbulb to turn on when you throw a switch.
So, you build the circuit if Figure 5. When the switch is connected to point A the capacitor
charges, and when you connect the switch to point B the light goes on. Let the resistance of
the light bulb be Rℓ = 1.50 ×103 Ω, the potential difference across the battery is VB = 10.0V ,
and the capacitor has a capacitance of C = 1.35 × 10−4 F .
(a) What should be the resistance of R1 , if we want the capacitor to charge to 90% of its
maximum charge in a time t = 0.500s
(b) How long from the time you start to discharge the capacitor, does it take for the current
through the light bulb to get to 33.3% of its initial current?
8. Scientist use a device called a mass spectrometer to identify charged sub-atomic particles.
They send the particles with a known velocity, into a device containing a known uniform
magnetic field, causing the particle to move in a circle as shown in Figure 6. By
Physics
Name
Institution
Answer question 1
Force between charges is computed through the use of Coulombs law which indicates that F=kqQ/r2. The constant k = 8.99*109Nm2/C2. Force is normally a vector . when more than one charge exerts force on another charge, the sum vector of such charge is the vector summation of the individual forces. Such charge is similar to gravitational forces between the interacting masses. In both cases, the force is computed as 1/r2. This implies that a force used to separate charges by 1/2r would be computed as (1/2r)2.
Answer to question 2
When a particle with charge Q=3.2*10-6c is stuck on a table and another with charge q= 7.7*10-7C and with mass m= 0.10 kg floating on the same article bearing charge Q, the distance of the article with charge q could be computed. The distance between the forces is given by the Coulombs law through the use of the formula F=kq1q2/r2.0.1newtons = 8.99*109*3.2*10-6*7.7*10-7/r2
R= 555.78
Answer to question 3
• Potential difference between the two plates is equal to velocity which is equal to 6.0*106m/s
• Force = mass *acceleration = 1.4*10-13*6.0*106 = -8254 nektons
The speed of the particles are computed by the formula V=ED. This is equal to 8.5*10-6*0.15. This is equal to 84.1
Answer to question 4
Voltage = current *resistance. This implies that in this case while V is 5.0 and resistance is 1.0*103, current will be equal to 5/1.0*103, = 500 amps
B the direction of the conventional current provides the electric charge movement from the positive side of the battery to its negative side as in indicated in the diagram below
Answer to question 5
• This section focuses on the equivalent resistance of a circuit. The equivalent resistance will be equal to (5.0*102+1.00*103)2. = 306.5.
• With voltage of 10, the currrent will be equal to voltage/resistance = 10/306.5= 0.03
Answer to question 6
• equivalent resistance of the ciruit = R1+R2/2 = 5.0*102+1.0*103=750
• The circuit is as indicated below
• The current = voltage/resistance = 10/750= 0.013
Answer to question in 7
•
E= V^2*C/2
Input:
Voltage across capacitor (V)
Capacitance (uF)
Load Resistance (Optional) (Ohms)
b.
Energy (Joules)
Time Constant (seconds)
Source: International Conference on Mechatronics Engineering and Electrical Engineering, & Sheng, (2015).
Answerr to question 8
• The detector should beplaced on the left side as the particle would be mopving in rounds towards the left side
• If the charge to mass ratio of the particle is q/m = 9.58*107C/KG and the velocity is 3.25*106,, the radius of the protons path would be caculated through the use of the formula as 9.58*107C/3.25*106 = 2.94769^13
International Conference on Mechatronics Engineering and Electrical Engineering, & In Sheng, A. (2015). Mechatronics engineering and electrical engineering: Proceedings of the 2014 International Conference on Mechatronics Engineering and Electrical Engineering (CMEEE 2014), Sanya, Hainan, P.R. China, 17-19 October 2014.
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