Anonymous asked in Science & MathematicsPhysics · 1 decade ago

can someone solve this physics problems.?

1 The mass of three wires of copper are in the ratio 1:3:5 and their lengths are in the ratio 5::1.Find the ratio of their electrical resistance.

2. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m if a force is applied at one end of the rope,the force which the rope exerts on the block will be what?

3. A wire has a resistance 30 ohm. it is bent in the form of a regular hexagon ABCDEF and the two ends are joined together . Find the effective resistance between the point A and B.

4. the area of a quadrilateral ABCD whose sides are 9m, 40m, 28m and 15m will be what.

5 Answers

  • KenK
    Lv 6
    1 decade ago
    Favourite answer

    3) Each leg of the hexagon is the same length, so each leg is 5 ohms. Since the wire is connected, there are 2 paths for current to flow 1) Directly from A to B, and 2) From B to C to D to E to F to A (5 sides). This acts like a 5 ohm resistor in parallel with a 25 ohm resistor. I will let you figure it from here, but I would guess the answer is approximately 4.2? ohms.

    #2) Since there is no friction, both objects use force to accelerate based on their mass. The total mass is m + M. The force = mass * acceleration equation is then written as f = (m + M) * a

    The force used to accelerate the box is the box's mass in a ration to the total , or M/(m + M).

    I hope that helps.

  • 1 decade ago

    I hope the lengths are in the ratio 5:3:1.

    The resistance is directly proportional to the length.

    The resistance is inversely proportional to the Area of cross section

    R = k * l/A

    If m is the mass, then its area of cross section A = m / {l*d} where d is the density and l is its length.

    R = k *l^2 d / m.

    d is a constant.

    R = Kl^2/m

    R1:R2:R3 = l1^2/m1:L2^2/m2:l3^2/m3 = 5^2/1: 3^2/3:1^2/5




    If a is the acceleration of the block, and If T is the tension,

    the force pulling the block is T = F- ma.

    If the block moves with constant velocity then a = 0 and

    T = F.



    The resistance of AB is R and that of AFEDCB is 5R.

    1/ R (effective) = 1/R + 1/ [5R] = [6/5] *[1/R] = 6/25 since R =5

    R effective = 4.17 ohm


    We often get questions about the area of parcels of land. We've created a tool (below) that should help you if your property has four sides you can measure. In addition to the four sides, we'll need to know either a diagonal or the degrees of one of the angles


  • 1 decade ago

    1. let, masses are m,3m,5m and lengths are 5l, 3l and, ratio of resistances r1:r2:r3=p(5l)/A1 : p(3l)/A2 : p(l)/A3 = 5pl^2/V1 : 3pl^2/V2 : pl^2/V3 = 5/(m/d) : 3/(3m/d) : 1/(5m/d) = 5 : 1 : (1/5)

    2. if force at the end of the rope is F, F=(M+m)a

    the force exerted on block is T=Ma=MF/(M+m)

    3. points A and B are joined parallely by part AB(5 ohm) and AFEDC(5+5+5+5+5=25 ohm). so, if equivalent resistance is R, then 1/R = 1/5 + 1/25.......R=4.16667 ohm

    4. you should have given required angles.

  • ?
    Lv 4
    4 years ago

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  • 1 decade ago

    do your homework yourself dude!!

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