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ziro4ka [17]
3 years ago
13

We’ve already investigated this problem with one spring scale in Think About 34.1. Now, imagine you have two spring scales, A an

d B, connected at the end of the scale that doesn’t move. The end that moves of each spring scale (where you take readings from) is attached to a string that goes over a pulley and connects to a 1 kg mass for both spring scales A and B. 1 kg 1 kg Spring Scale A Spring Scale B (a) State what you think each spring scale will read in this situation. (b) Construct a logical argument that explains why the spring scales read
Physics
1 answer:
Debora [2.8K]3 years ago
5 0

Answer:

Explained

Explanation:

1.Each of the spring scale will read 10N,considering acceleration due to gravity as 10 m/s^2

2.Each of the spring scale will read 10N because each string exerts a force of 10 N to counterbalance the force of 1 kg mass attached to it. This means the tension on the both side of the string is 10 N. So the scale will read 10 N. Also as spring balances are attached in series and kept on table so both spring balances will read same readings.

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Two circular coils are concentric and lie in the same plane. The inner coil contains 110 turns of wire, has a radius of 0.014 m,
Hatshy [7]

Answer:

The current flowing through the outer coils is  

Explanation:

From the question we are told that

   The number of turn of inner coil is N _i  =  110 \  turns

    The radius of inner coil is  r_i  =  0.014 \ m

     The current flowing through the inner coil is  I_i  =  9.0 \ A

     The number of turn of outer coil is N_o  =  160 \ turns

     The radius of outer  coil is r_o  =  0.022\ m

For net magnetic field at the common center of the two coils to be  zero  the current flowing in the outer coil must be opposite to current flowing inner coil

   The magnetic field due to inner coils  is mathematically represented as

            B_i  =   \frac{N_i \mu I}{2 r_i}

     The magnetic field due to inner coils  is mathematically represented as

            B_o  =  \frac{N_o \mu I_o}{2 r_o}

Now for magnetic field at center to be zero

             B_o  =  B_i

So

         \frac{N_i \mu I_i}{2 r_i} =  \frac{N_o \mu I_o}{2 r_o}

=>      \frac{110 * 9}{2 *  0.014} =  \frac{160 *I_o}{2 0.022}

         I_o  = 9.72 \ A

6 0
4 years ago
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