K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2
So, option B is your answer!
Hope this helps!
Answer:
you would have to stand 6 ft back
Explanation:
<h2>
Hey There!</h2><h2>
_____________________________________</h2><h2>
Answer:</h2><h2 /><h2>
</h2><h2>
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<h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
<h2>
SOLUTION:</h2>
According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,
Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,
W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>
Answer:
1) The charge on the outer shell is +4·Q
2) The charge on the inner shell is +Q
Explanation:
1) The given parameters of the spherical shell are;
The net charge on the spherical shell = 3·Q
The point charge surrounded by the spherical shell = -Q
Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;
The net charge, 3·Q = -q + x
∴ x = 3·Q + Q = 4·Q
The charge on the outer shell, x = 4·Q
2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;
-Q + y = 0
∴ y = +Q
The charge on the inner shell, y = +Q