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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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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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  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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Question Bank
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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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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 Bank
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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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Heman
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HemanLEVEL 18
Asked: August 19, 2020In: Physics

A 550 N physics student stands on a bathroom scale in an elevator . As the elevator starts moving the scale reads 450N .Find the magnitude and direction of the acceleration of the elevator .

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  1. Tmg enoch LEVEL 14
    Added an answer on August 20, 2020 at 11:46 am

    Solution, Given, The true weight of student (w) =550 N Apparent weight (R)=450 N The magnitude and direction of elevator =? Here, the true weight of body acts downwards and apparent weight (the reaction of elevator ) is upward . so , free body diagram is , Net force (f)  ma =W-R or , (550/9.8) a = 5Read more

    Solution,
    Given,
    The true weight of student (w) =550 N
    Apparent weight (R)=450 N
    The magnitude and direction of elevator =?
    Here, the true weight of body acts downwards and apparent weight (the reaction of elevator ) is upward . so , free body diagram is ,
    Net force (f) 
    ma =W-R
    or , (550/9.8) a = 550-450 =100
    or , a = 100×9.8/550
    ∴ a =1.78m/s2 downwards

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

A mass of 1 kg is attached to the lower end of a string 1 m long whose upper end is fixed . The mass is made to rotate in a horizontal circle of radius 60 cm . if the circular speed of the mass is constant , find the tension in the string and the period of motion .

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Answer
  1. Heman LEVEL 18
    Added an answer on August 27, 2020 at 12:28 pm

    Given , Mass of object (m) = 1kg  Length of string (l) = 4m Radius of circle (r) = 60cm Tension on the string (T) = ? Period of motion = ? From figure , Tsinθ = mv2/r Tcosθ =mg tanθ = v2/rg and  sinθ = r/l = 0.6/1 = 0.6 or , θ = sin-1(0.6) = 36.87 so from relation  Tcosθ =mg T = mg/cosθ =1x10/cos36.Read more

    Given ,
    Mass of object (m) = 1kg 
    Length of string (l) = 4m
    Radius of circle (r) = 60cm
    Tension on the string (T) = ?
    Period of motion = ?
    From figure ,
    Tsinθ = mv2/r
    Tcosθ =mg
    tanθ = v2/rg and 
    sinθ = r/l = 0.6/1 = 0.6
    or , θ = sin-1(0.6) = 36.87
    so from relation 
    Tcosθ =mg
    T = mg/cosθ =1×10/cos36.87 = 12.5 N
    Again , from relation 
    tanθ = v2/rg
    or , t = 2π√lcosθ/g
             = 2π√1xcos36.87/10 = 1.78s

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knowithere
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knowithereLEVEL 1
Asked: June 12, 2020In: Physics

A pendulum clock is in an elevator that descends at constant velocity. Does it keep correct time? If the some clock is in an elevator in free fall, does it keep correct time? 

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  1. Sagar poudel LEVEL 1
    Added an answer on June 21, 2020 at 8:38 am

    The time period of a pendulum of length is given by T=2pi squareroot l/g. In an elevator that descends at constant velocity , the acceleration due to gravity does not altered and hence, it gives the correct time. When the elevator in free fall, the effective value of acceleration due to gravity g isRead more

    The time period of a pendulum of length is given by T=2pi squareroot l/g. In an elevator that descends at constant velocity , the acceleration due to gravity does not altered and hence, it gives the correct time. When the elevator in free fall, the effective value of acceleration due to gravity g is zero and hence time period is infinite . Such a pendulum will be at rest and do not oscillate when displaced from their mean positions . So, it cannot keep time, no possibility of correct time.

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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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