A ) decrease.
B ) increase.
C ) increase, then decrease.
D ) not change.
The answer is A) decrease
Take pushing a box, for example-- You push your hardest then give out, still trying to push the box. You are doing less work than what you have started with!
( Mind marking me for branliest? ; ) )
As we know that as per Newton's II law we have

here we will have
= change in momentum
= time interval in which momentum is changed
now in order to have least injury during jumping we need to have least force on the jumper
so in order to have least force we can say that the momentum must have to change in maximum time so that amount of force must be least
So we need to increase the time in which momentum of the system is changed
Answer:
c. because A will land first becuase its heavier :)
Explanation:
Answer:
magnitude=34.45 m
direction=
Explanation:
Assuming the initial point P1 of this vector is at the origin:
P1=(X1,Y1)=(0,0)
And knowing the other point is P2=(X2,Y2)=(19.5,28.4)
We can find the magnitude and direction of this vector, taking into account a vector has a initial and a final point, with an x-component and a y-component.
For the magnitude we will use the formula to calculate the distance
between two points:
(1)
(2)
(3)
(4) This is the magnitude of the vector
For the direction, which is the measure of the angle the vector makes with a horizontal line, we will use the following formula:
(5)
(6)
(7)
Finding
:
(8)
(9) This is the direction of the vector
Answer:
(a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.
Explanation:
Given that,
Time = 1.95 s
Distance = 23.5 m
(a). We need to calculate the velocity
Using equation of motion

Put the value into the formula



(b). We need to calculate the total distance the object travels during the fall
Using equation of motion

Put the value in the equation



The total time is


We need to calculate the distance
Using equation of motion

Put the value into the formula


Hence, (a). The velocity of the object is -2.496 m/s.
(b). The total distance of the object travels during the fall is 23.80 m.