Explanation:
Given that,
A student travels 11 m north and then turns around to travel 25 m south.
Total time, t = 12 s
The total distance or the total path covered by the student is equal to 11 m + 25 m = 36 m
Displacement of the student or the shortest path covered is d = 25-11 = 14 m
(a) The student's average speed = total distance/total time
(b) The student's average velocity = total displacement/total time
Answer:
a) p = m1 v1 + m2 v2
, b) dp / dt = m1 a1 + m2 a2
, c) It is equivalent to force
dp / dt = 0
Explanation:
In this problem we have two blocks and the system is formed by the two bodies.
Part A. Initially they ask us to find the moment of the whole system
p = m1 v1 + m2 v2
Part B.
Find the derivative
dp / dt = m1 dv1dt + m2 dv2 / dt
dp / dt = m1 a1 + m2 a2
Part C.
Let's analyze the dimensions
m a = [kg] [m / s2] = [N]
It is equivalent to force
Part d
Acceleration is due to a net force applied
Part e
The acceleration of block 1 is due to the force exerted by block 2 during the moment change
Part f
Force of block 1 on block 2
True f12 = m1a1 f21 = m2a2
Part g
By the law of action and reaction are equal magnitude F12 = f21
Part H
dp / dt = 0
Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed
Answer:
0.00354 (N)
Explanation:
Convert to metric system:
Formula for gravitational force:
where s is the distance between 2 bodies masses m and M
Substitute the number to the formula above and since the 2 forces are acting in opposite direction, the total net gravitational force on the mass of origin be:
Answer:
(a) 5.7 s
(b) 39 m/s
Explanation:
(a) u = 18 m/s
At the maximum height, the final velocity of ball is zero. lte teh time taken by the ball to go from 50 m height to maximum height is t.
use first equation of motion.
v = u + g t
0 = 18 - 10 x t
t = 1.8 s
Let the maximum height attained by the ball when it thrown from 50 m height is h'.
Use third equation of motion
v^2 = u^2 + 2 g h'
0 = 18^2 - 2 x 10 x h'
h' = 16.2 m
Total height from the ground H = h + h' = 50 + 16.2 = 76.2 m
Let t' be the time taken by the ball to hit the ground as it falls from maximum height.
use third equation of motion
H = ut + 1/2 x g t'^2
76.2 = 0 + 1/2 x 10 x t'^2
t' = 3.9 s
Total time taken by the ball to hit the ground = T = t + t' = 1.8 + 3.9 = 5.7 s
(b) Let v be the velocity with which the ball strikes the ground.
v^2 = u^2 + 2 g H
v^2 = 0 + 2 x 10 x 76.2
v = 39 m/s
