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LuckyWell [14K]
3 years ago
7

Plz help

Physics
1 answer:
katrin2010 [14]3 years ago
4 0

The particles of the medium (slinky in this case) move up and down (choice #2) in a transverse wave scenario.

This is the defining characteristic of transverse waves, like particles on the surface of water while a wave travels on it, or like particles in a slack rope when someone sends a wave through by giving it a jolt.

The other kind of waves is longitudinal, where the particles of the medium move "left-and-right" along the direction of the wave propagation. In the case of the slinky, this would be achieved by giving a tensioned slinky an "inward" jolt. You would see that such a jolt would give rise to a longitudinal wave traveling along the length of the tensioned slinky. Another example of longitudinal waves are sound waves.

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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
3 years ago
A spring with an 8-kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equili
pentagon [3]

Answer:

x = 1.26 sin 3.16 t

Explanation:

Assume that the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

V=\dfrac{dx}{dt}

V= A ω cosωt

Maximum velocity

V(max)= Aω

Given that F= 32 N

F = K Δ

K=Spring constant

Δ = 0.4 m

32 =0.4 K

K = 80 N/m

We know that  ω²m = K

8 ω² = 80

ω = 3.16 s⁻¹

Given that V(max)= Aω = 4 m/s

3.16 A = 4

A= 1.26 m

Therefore the general equation of displacement

x = 1.26 sin 3.16 t

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