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Tmg enoch
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Tmg enochLEVEL 14
Asked: August 22, 2020In: Physics

Calculate the value of two equal charges if they repel one another with a force of 0.1 N when situated 50 cm apart in vaccum . what would be the distance between them if they are placed in an insulating medium of dielectric constant 10?

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  1. Heman LEVEL 18
    Added an answer on August 23, 2020 at 9:40 pm

    Given, Force (f) =0.1 N Distance between Charge (r) =540 cm  Charge (q) =? we know that , F = 1/4π∈o.qxq/x2 ( air ) or , 0.1 =9x109xq2/(0.5)2 or , √0.1x0.25/9x109 = q  or ,q =√0.025/9x109         =√0.000278x10-8        =√2.78x10-12 ∴ q =1.67x10-6 c In the second part , a medium of dielectric constanRead more

    Given,
    Force (f) =0.1 N
    Distance between Charge (r) =540 cm 
    Charge (q) =?
    we know that ,
    F = 1/4π∈o.qxq/x2 ( air )
    or , 0.1 =9×109xq2/(0.5)2
    or , √0.1×0.25/9×109 = q 
    or ,q =√0.025/9×109
            =√0.000278×10-8
           =√2.78×10-12
    ∴ q =1.67×10-6 c
    In the second part , a medium of dielectric constant is placed i.e. k =10 , Distance (r) =?
    F =1/4π∈ok .q2/r2
    or , 0.1 =9×109/10 x(1.67×10-6)2/r2
    or , r =√9×109x(1.67×10-6)2
    or , r =√25.10×10-3
    ∴r =0.16 m

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Question BankLEVEL 6
Asked: August 31, 2020In: Physics

Two insulated charged copper spheres A and B of identical size have charges qA and qB respectively. A third sphere C of the same size but uncharged is brought in contact with the first and then in contact with the second and finally removed from both. What are the new charges on A and B1

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  1. knowithere LEVEL 1
    Added an answer on September 12, 2020 at 8:23 am

    A the third body which is uncharged contact with body A then both distribute net charge equally. e.g. Qnet=(0+qa) because they equally distribute so, both have finally qa/2 charge available. now third body contact with body B then Qnet=(qa/2+qb) now both distribute charge equally hence, each has (qaRead more

    A the third body which is uncharged contact with body A then both distribute net charge equally.
    e.g. Qnet=(0+qa)
    because they equally distribute so, both have finally qa/2 charge available.
    now third body contact with body B
    then Qnet=(qa/2+qb)
    now both distribute charge equally
    hence,
    each has (qa/2+qb)/2

    hence finally charge in body A=qa/2
    charge in body B=(qa/2+qb)/2=qa/4+qb/2

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Question BankLEVEL 1
Asked: August 14, 2020In: Physics MCQs

A string is tied on a sonometer, second end is hanging downward through a pulley with tension T. The velocity of the transverse wave produced is proportional to

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a) 1/√T b) √T c) T d) 1/T

a) 1/√T

b) √T

c) T

d) 1/T

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Answer
  1. knowithere LEVEL 21
    Added an answer on August 18, 2020 at 12:06 pm

    Option b is the correct answer!!! Explanation: v=√(T/m) v∝√T.

    Option b is the correct answer!!!
    Explanation: v=√(T/m)
    v∝√T.

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Tmg enoch
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Tmg enochLEVEL 14
Asked: August 20, 2020In: Physics

A body of mass 200 g is executing simple harmonic motion with amplitude of 20 mm . The maximum force which acts upon it is 0.8 N . calculate its maximum velocity and its period of oscillation .

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Answer
  1. Heman LEVEL 18
    Added an answer on August 27, 2020 at 11:01 am

    Given , Mass of the body (m) = 200g  Amplitude (A) = 20mm Maximum force (F) = 0.8 N Maximum velocity (vmax) =? Period of oscillation (T) =? We know that Fmax = mv2max/A or , 0.8 = 200x10-3xv2max/20x10-3 or , 0.8x20/200= v2max or , vmax = 0.28 m/s Again, vmax = ωA or , v = 2π/T . A  or , T = 2πr/v =2Read more

    Given ,
    Mass of the body (m) = 200g 
    Amplitude (A) = 20mm
    Maximum force (F) = 0.8 N
    Maximum velocity (vmax) =?
    Period of oscillation (T) =?
    We know that
    Fmax = mv2max/A
    or , 0.8 = 200×10-3xv2max/20×10-3
    or , 0.8×20/200= v2max
    or , vmax = 0.28 m/s
    Again,
    vmax = ωA
    or , v = 2π/T . A 
    or , T = 2πr/v =2πx20x10-3/0.28 =0.45s
    ∴ Maximum velocity = 0.28 m/s2
    Time period =0.45s

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Kabin 2.0
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Kabin 2.0
Asked: June 20, 2020In: Physics MCQs

Joule-Ohm/Second-volt is equal to

  • 4

a. Ampere b. Tesla c. Watt d. Volt

a. Ampere

b. Tesla

c. Watt

d. Volt

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Answer
  1. Kabindra Dhakal
    Added an answer on June 20, 2020 at 10:49 pm

    The correct answer is option (d).

    The correct answer is option (d).

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Question BankLEVEL 1
Asked: September 5, 2020In: Computer MCQs

Which of the following is not the External Security Threats?

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A. Front-door Threats B. Back-door Threats C. Underground Threats D. Denial of Service (DoS)

A. Front-door Threats
B. Back-door Threats
C. Underground Threats
D. Denial of Service (DoS)

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Answer
  1. knowithere LEVEL 21
    Added an answer on September 9, 2020 at 4:31 pm

    Option C is the correct answer!!!  

    Option C is the correct answer!!!

     

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Question BankLEVEL 1
Asked: September 5, 2020In: Computer MCQs

Which of the following is correct regarding Class B Address of IP address

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A. Network bit – 14, Host bit – 16 B. Network bit – 16, Host bit – 14 C. Network bit – 18, Host bit – 16 D. Network bit – 12, Host bit – 14

A. Network bit – 14, Host bit – 16
B. Network bit – 16, Host bit – 14
C. Network bit – 18, Host bit – 16
D. Network bit – 12, Host bit – 14

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Answer
  1. knowithere LEVEL 21
    Added an answer on September 10, 2020 at 12:29 pm

    Option A is the correct answer!!! Network bit – 14, Host bit – 16

    Option A is the correct answer!!!

    Network bit – 14, Host bit – 16

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Tmg enoch
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Tmg enochLEVEL 14
Asked: August 22, 2020In: Physics

A charge oil drop remains stationary when situated between two parallel horizontal metal plates 25 mm apart and a p.d. of 1000v is applied to the plates . find the charge on the drop if it has mass of 5 x10^-15 kg.

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Answer
  1. Heman LEVEL 18
    Added an answer on August 23, 2020 at 9:17 pm

    Given, Separation of plates (d) =25mm Mass of drop (m) =5x10-15kg Potential difference (V) =1000v Charge on the drop (q) =? if the oil drop remains stationary , then  The force on oil drop due to electric field =weight of the drop  i.e. qE=mg or ,q =mg/E =mg/v/d =mgd/v         = 5x10-15x10x25x10-3/1Read more

    Given,
    Separation of plates (d) =25mm
    Mass of drop (m) =5×10-15kg
    Potential difference (V) =1000v
    Charge on the drop (q) =?
    if the oil drop remains stationary , then 
    The force on oil drop due to electric field =weight of the drop 
    i.e. qE=mg
    or ,q =mg/E =mg/v/d =mgd/v
            = 5×10-15x10x25x10-3/1000 =1.25×10-18c
    ∴The charge on the oil drop q =1.25×10-18c
    Hence the required charge on the oil drop is 1.25×10-18c.

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Tmg enoch
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Tmg enochLEVEL 14
Asked: August 20, 2020In: Physics

A 0.15 kg glider is moving to the right on a frictionless horizontal air track with a speed of 0.80 m/s . It has a head on collision with a 0.300 kg glider that is moving to the left with a speed of 2.2 m/s . Find the final velocity ( magnitude and direction ) of each glider if the glider if the collision is elastic .

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Answer
  1. Heman LEVEL 18
    Added an answer on August 27, 2020 at 1:05 pm

    Given , Mass of glider A (M1) = 0.15 kg  Mass of glider B (M2) = 0.3 kg Initial velocity of A (u1) = 0.80ms-1 Initial velocity of B ( u2) = -2.2m/s Final velocity of A (v1) =? Final velocity of B (v2) = ? we have, M1u1+M2u2 = M1v1+M2v2 0.15x0.8+0.3x(-2.2) = 0.15v1+0.3v2 or , -0.54 = 0.15v1+0.3v2 ...Read more

    Given ,
    Mass of glider A (M1) = 0.15 kg 
    Mass of glider B (M2) = 0.3 kg
    Initial velocity of A (u1) = 0.80ms-1
    Initial velocity of B ( u2) = -2.2m/s
    Final velocity of A (v1) =?
    Final velocity of B (v2) = ?
    we have,
    M1u1+M2u2 = M1v1+M2v2
    0.15×0.8+0.3x(-2.2) = 0.15v1+0.3v2
    or , -0.54 = 0.15v1+0.3v2 ………i
    Again for elastic collision,
    u1+v1 =u2+v2
    or , 0.8 +v1 =-2.2+v2
    or , v1-v2 = -3
    Solving (i) and (ii) , we get 
    v2 =-0.2m/s
    and v1 = -3.2 m/s

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Question Bank
Asked: January 15, 2021In: Biology MCQs

Which of the following is a nucleotide found in DNA?

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A) Deoxyribose + Phosphate Group + Polymerase B) Adenine + Phosphate Group + Thymine C) Deoxyribose + Phosphate Group + Thymine D) Cytosine + Phosphate Group + Uracil

A) Deoxyribose + Phosphate Group + Polymerase
B) Adenine + Phosphate Group + Thymine
C) Deoxyribose + Phosphate Group + Thymine
D) Cytosine + Phosphate Group + Uracil

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Answer
  1. knowithere LEVEL 21
    Added an answer on January 22, 2021 at 8:02 pm

    Option C is the correct answer!!! Deoxyribose + Phosphate Group + Thymine is the only option in that form of nucleotide in DNA.

    Option C is the correct answer!!!

    Deoxyribose + Phosphate Group + Thymine is the only option in that form of nucleotide in DNA.

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