The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
pls thank me and brainliest me
Answer:
Changes in the object's momentum (answer D)
Explanation:
A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.
So, select the last option (D) in the given list.
Answer:
m = 14*26 = 364
Explanation:
overall magnification is given as m

mo magnification of objective lens
me magnification of EYE lens
where mo is given as

and me as

d is distant of distinct vision = 25.0 cm for normal eye
fe = focal length of eye piece
focal length of objective lense is 0.140 cm
we know that








m = 14*26 = 364
Hello
Here we must use the equation of motion
v^2 = u^2 + 2as; where v is final velocity, u is initial velocity, a is the acceleratoin and is the distance travelled.
We select this one because the time of collision is unknown to us.
We know the truck stopped so its final velocity is 0; thus v = 0.
Converting the initial velocity to SI units, we get 3.89 m/s.
The distance traveled, s, is 0.062 meters.
Inserting all of these values into the equation,
0 = (3.89)^2 + 2(a)(0.062)
and solving for a, we get a to be
-122.0 ms^(-2)
The negative sign indicates the acceleration is in the opposite direction to the initial motion, which means the truck decelerated. This is consistent with the given condition.