Answer:
x = 10.53 m
Explanation:
Let's analyze this problem a bit, the time that the cowboy must take to fall must be the time that the horse takes to arrive
Let's start by looking for the cowboy's time, which starts from rest and the point where the chair is is y = 0
y =y₀ + v₀ t - ½ g t²
0 = y₀ - ½ g t²
t = 
we calculate
t = √(2 3.22 / 9.8)
t = 0.81 s
the horse goes at a constant speed
x = 
t
x = 13 0.81
x = 10.53 m
this is the distance where the horse should be when in cowboy it is left Cartesian
The force of gravity is that amount of attractive force which is exerted between two bodies on each other.
As per Newton's law of gravitation the force of gravity is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of separation distance between them.
Mathematically 
Here G is the gravitational constant whose value is 6.67×
As per the question the masses of the objects are increased keeping the separation distance same.
As force of gravity ∝ product of masses of two bodies,
Hence the force of gravity increases with the increase of mass.
The answer to your question is True.
Hope this helped...even though your question has been answered seven days later.
Better late than never xD
The vertical forces add up to zero.
The net force on the object is 60N to the right.
F = m • a
60N right = (10kg) • (a)
a = (60N right) / (10kg)
a = 6 m/s^2 to the right
Multi-engine Commercial would be nice.
<span>Air resistance will eventually equal the thrust force, so there is no resultant force meaning the car will stay at the same speed and can no longer accelerate. </span>
