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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}

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Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

Mass of the truck = 3047.8 kg

The ramp has an angle of 9.5 °

Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

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<u>Step 3: </u>Calculate potential energy

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5 0
3 years ago
Two cars are traveling along a straight line in the same direction, the lead car at 25 m/s and the other car at 35 m/s. At the m
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Answer:

a. t_1=12.5\ s

b. a_2=-13.61\ m.s^{-2}  must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c. t_2=2.5714\ s is the time taken to stop after braking

Explanation:

Given:

  • speed of leading car, u_1=25\ m.s^{-1}
  • speed of lagging car, u_{2}=35\ m.s^{-1}
  • distance between the cars, \Delta s=45\ m
  • deceleration of the leading car after braking, a_1=-2\ m.s^{-2}

a.

Time taken by the car to stop:

v_1=u_1+a_1.t_1

where:

v_1=0 , final velocity after braking

t_1= time taken

0=25-2\times t_1

t_1=12.5\ s

b.

using the eq. of motion for the given condition:

v_2^2=u_2^2+2.a_2.\Delta s

where:

v_2= final velocity of the chasing car after braking = 0

a_2= acceleration of the chasing car after braking

0^2=35^2+2\times a_2\times 45

a_2=-13.61\ m.s^{-2} must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

c.

time taken by the chasing car to stop:

v_2=u_2+a_2.t_2

0=35-13.61\times t_2

t_2=2.5714\ s  is the time taken to stop after braking

7 0
3 years ago
SOMEONE PLEASE HELP PLEASE PLEASE
Black_prince [1.1K]
Ais the correct answer
4 0
2 years ago
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