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YOUR ANSWER IS <em><u>D.PLACE</u></em><em><u> </u></em><em><u>IT</u></em><em><u> </u></em><em><u>IN</u></em><em><u> </u></em><em><u>AN</u></em><em><u> </u></em><em><u>ELECTRI</u></em><em><u>C</u></em><em><u> </u></em><em><u>FIELD</u></em>
<em><u>BE</u></em><em><u>CAUSE</u></em><em><u> </u></em><em><u>IT </u></em><em><u>makes</u></em><em><u> sense you can use alternating current to remove magnetism</u></em>
2) transverse
hope this helped:)
<h2>
Speed with which it return to its initial level is 100 m/s</h2>
Explanation:
We have equation of motion v² = u² + 2as
Initial velocity, u = 100 m/s
Acceleration, a = -9.81 m/s²
Final velocity, v = ?
Displacement, s = 0 m
Substituting
v² = u² + 2as
v² = 100² + 2 x -9.81 x 0
v² = 100²
v = ±100 m/s
+100 m/s is initial velocity and -100 m/s is final velocity.
Speed with which it return to its initial level is 100 m/s
Answer:
K = 588.3 N/m
Explanation:
From a forces diagram, and knowing that for the maximum value of K, the crate will try to rebound back up (Friction force will point downward):
Fe - Ff - W*sin(22) = 0 Replacing Fe = K*X and then solving for X:

By conservation of energy:

Replacing our previous value for X and solving the equation for K, we get maximum value to prevent the crate from rebound:
K = 588.3 N/m