1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
sesenic [268]
3 years ago
13

Four identical metallic spheres with charges of 2.2 µC, 4.2 µC, −7.4 µC, and −6.8 µC are placed on a piece of paper. The paper i

s lifted on all corners so that the spheres come into contact with each other simultaneously. The paper is then flattened so that the metallic spheres become separated.
(a) What is the resulting charge on each sphere?
(b) How many excess or absent electrons (depending on the sign of your answer to part (a)) correspond to the resulting charge on each sphere?
Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

Part a)

charge on each sphere is -1.95 micro coulomb

Part b)

For first sphere

N = 2.6 \times 10^{13} excess charge

For second sphere

N = 3.8 \times 10^{13} excess charge

For third sphere

N = 3.4 \times 10^{13} absent charge

For third sphere

N = 3.03 \times 10^{13} absent charge

Explanation:

Part a)

Since all the spheres are of identical size so the total charge of the sphere will divide equally on them

So we have

q = \frac{Q_1 + Q_2 + Q_3 + Q_4}{4}

q = \frac{2.2 \mu C + 4.2 \mu C - 7.4 \mu C - 6.8 \mu C}{4}

q = -1.95 \mu C

So charge on each sphere is -1.95 micro coulomb

Part b)

For first sphere

initial charge = 2.2 micro coulomb

final charge = -1.95 micro coulomb

excess charge = -1.95 - 2.2 = -4.15 micro coulomb

Q = Ne

N = \frac{4.15\times 10^{-19}}{1.6 \times 10^{-19}}

N = 2.6 \times 10^{13}

For second sphere

initial charge = 4.2 micro coulomb

final charge = -1.95 micro coulomb

excess charge = -1.95 - 4.2 = -6.15 micro coulomb

Q = Ne

N = \frac{6.15\times 10^{-19}}{1.6 \times 10^{-19}}

N = 3.8 \times 10^{13}

For third sphere

initial charge = -7.4 micro coulomb

final charge = -1.95 micro coulomb

absent charge = -1.95 + 7.4 = 5.45 micro coulomb

Q = Ne

N = \frac{5.45\times 10^{-19}}{1.6 \times 10^{-19}}

N = 3.4 \times 10^{13}

For third sphere

initial charge = -6.8 micro coulomb

final charge = -1.95 micro coulomb

absent charge = -1.95 + 6.8 = 4.85 micro coulomb

Q = Ne

N = \frac{4.85\times 10^{-19}}{1.6 \times 10^{-19}}

N = 3.03 \times 10^{13}

You might be interested in
6. Which does more wor:
umka2103 [35]

Explanation:

1N=9,81 kg

1. 9,81×500×10=<u>49.050</u>

2. 9,81×100×40=39.240

option 1 does more wor

8 0
2 years ago
1 poi
tatuchka [14]

Answer:

true

Explanation:

for example assume you are setting in a moving bus and when someone see you from the ground you are in motion but for some who is with you in the bus you are not in motion.

6 0
2 years ago
A 150 g sample of brass at 100 °C is placed in a Styrofoam cup of water containing 120 mL of water at 10 °C. No heat is lost to
fredd [130]

Answer:

≈19.144°C.

Explanation:

all the details are in the attachment.

Note, that c₁, m₁, t₁ are the parameters of the sample of brass; c₂, m₂ and t₂ are  the parameters of the sample of water.

P.S. change the provided design according Your requirements.

4 0
2 years ago
Suppose an astronaut has landed on Planet * Fully equipped, the astronaut has a
gayaneshka [121]

From the calculations, the value of the acceleration due to gravity is 0.38 m/s^2.

<h3>What is weight?</h3>

The weight of an object is obtained as the product of the mass of the body and the acceleration due to gravity.

Thus;

When;

mass = 120 kg

weight =  46 N

acceleration due to gravity = 46 N/120 kg

=0.38 m/s^2

Learn more about acceleration due to gravity :brainly.com/question/13860566

#SPJ1

7 0
1 year ago
In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
Other questions:
  • Two railroad cars, each of mass 6500 kg and traveling 80 km/h in opposite directions, collide head-on and come to rest. how much
    10·1 answer
  • Help me <br><br> If you answered this, you will have 30 points :)
    9·1 answer
  • Why do you think the earth has fewer craters than the moon
    8·1 answer
  • Perform an Internet search to learn about grounding wires, fuses, and circuit breakers. Specifically,
    11·1 answer
  • The largest stars are _____ times the mass of the Sun. 100 10,000 10 1,000The largest stars are _____ times the mass of the Sun.
    11·1 answer
  • A man weighing 490N on earth weighs 81.7N on the moon.His mass on the moon is kg
    15·2 answers
  • You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it
    6·1 answer
  • Help me out with this stuff
    10·2 answers
  • Joel uses a force of 50 Newtons to hold two weights 0.60 meters above his head. How much work is Joel doing on the weights?
    12·1 answer
  • Which is the most accurate description of how the coolant works in an engine?
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!