Answer:
1. 
2. 
3. 
Explanation:
Given:
- mass of slinky,

- length of slinky,

- amplitude of wave pulse,

- time taken by the wave pulse to travel down the length,

- frequency of wave pulse,

1.



2.
<em>Now, we find the linear mass density of the slinky.</em>


We have the relation involving the tension force as:




3.
We have the relation for wavelength as:



Answer:
2*10^-<em>5</em>
Explanation:
<em>B=</em><em>I</em><em>L</em>
<em>I=</em><em>B</em><em>/</em><em>L</em>
<em>I=</em><em>0</em><em>.</em><em>0</em><em>0</em><em>2</em><em>0</em><em>*</em><em>1</em><em>0</em><em>^</em><em>-</em><em>4</em><em>/</em><em>1</em><em>0</em>
<em>I=</em><em>2</em><em>*</em><em>1</em><em>0</em><em>^</em><em>5</em>
30 Grams would be your answer (I took the test and got it right)