There was no net force on the stuffed toy, because the kids might have the same strength, The same force is on both sides of it. T<span>hey cancel each other out. They exert a force on the stuffed toy equal in strength but opposite in direction. The forces are balanced and the stuffed toy does not move. </span>Its like a game of tug-o-war, but you and I have the same strength. the rope would be still and not moving.
Answer:
The size of the force that pushes the wall is 12,250 N.
Explanation:
Given;
mass of the wrecking ball, m = 1500 kg
speed of the wrecking ball, v = 3.5 m/s
distance the ball moved the wall, d = 75 cm = 0.75 m
Apply the principle of work-energy theorem;
Kinetic energy of the wrecking ball = work done by the ball on the wall
¹/₂mv² = F x d
where;
F is the size of the force that pushes the wall
¹/₂mv² = F x d
¹/₂ x 1500 x 3.5² = F x 0.75
9187.5 = 0.75F
F = 9187.5 / 0.75
F = 12,250 N
Therefore, the size of the force that pushes the wall is 12,250 N.
Answer:
upward force acting = 261.6 N
Explanation:
given,
mass of gibbon = 9.4 kg
arm length = 0.6 m
speed of the swing
net force must provide

force of gravity = - mg

= 
= 
=9 x 29.067
= 261.6 N
upward force acting = 261.6 N
Answer:
The final velocity of the object is 330 m/s.
Explanation:
To solve this problem, we first must find the acceleration of the object. We can do this using Newton's Second Law, given by the following equation:
F = ma
If we plug in the values that we are given in the problem, we get:
42 = 7 (a)
To solve for a, we simply divide both sides of the equation by 7.
42/7 = 7a/7
a = 6 m/s^2
Next, we should write out all of the information we have and what we are looking for.
a = 6 m/s^2
v1 = 0 m/s
t = 55 s
v2 = ?
We can use a kinematic equation to solve this problem. We should use:
v2 = v1 + at
If we plug in the values listed above, we should get:
v2 = 0 + (6)(55)
Next, we should solve the problem by performing the multiplication on the right side of the equation.
v2 = 330 m/s
Therefore, the final velocity reached by the object is 330 m/s.
Hope this helps!