Answer:
(A) –14m/s
(B) –42.0m
Explanation:
The complete solution can be found in the attachment below.
This involves the knowledge of motion under the action of gravity.
Check below for the full solution to the problem.
Answer:
a) wet marble , dry marble, newspaper, and towel
(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
Answer:
b.only when the current in the first coil changes.
Explanation:
An induced current flow in the second coil only when there is a change in current in the first cool. A steady current will produce no change in flux (due to magnetic effect of a current) by the first coil, and according to Faraday, induced current is only produced when there is a change in flux linkage.
