Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
Answer:
induced current
Explanation:
intentionally manipulated.
Answer:
The minimum compression is 
Explanation:
From the question we are told that
The mass of the block is 
The spring constant is 
The coefficient of static friction is 
For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis
Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

where x is the minimum compression to keep the block from slipping
Now equating this two formulas and making x the subject

substituting values we have


<span>On the y-axis (the bottom of the table) hope this helps</span>
First,

where
is density,
is mass, and
is volume. We can compute the volume of the roll:


When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness
. Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).
So we have

where
is the given area, so


If we're taking significant digits into account, the volume we found would have been
, in turn making the thickness
.