22: A 2.0 kg object moves along the x-axis under the influence of an unknown force. Its potential energy due to the force is U(x) = ax4 + bx2 , where a = 2.5 J/m4 and b = 12 J/m2 a) What is the force on the object when x = 0.5 m? b) If the object is released from rest at x = 0.5 m, what is its velocity when it crosses the origin?

a) -13.25 N, b) 1.78 m/sec

Step-by-step explanation:

Since we have given that

where

As we know that

Here, x= 0.5 m

So, it becomes,

Now, we need to find the velocity when it crosses the origin,

Hence, a) -13.25 N, b) 1.78 m/sec

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

51

step-by-step explanation:

73-22=51 the   mass number minus the amount of protons gives you the amount of neutrons