Force required to accelerate 10 kg object to 5.9 m/s/s ?
Mass = 10 kg
Acceleration = 5.9 m/s^2
Force = Mass * Acceleration
Force = 10 kg * 5.9 m/s^2
Force = 59 kg m /s^2 = 59 N
Answer: Rn :)))) no explanation needed
An applied force<span> is a </span>force<span> that is </span>applied<span> to an object by a person or another object.
An attractive force is a force of an attraction (where object are attracted by each other). Gravitation is an example of attractive force.
</span>Normal force<span> is the component, perpendicular to the surface (surface being a plane) of contact.
</span><span>The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences attractive force from the air it passes through. It also experiences a downward pull because of the normal force.
Solution A.</span>
Answer:
Work done = 35467.278 J
Explanation:
Given:
Height of the cone = 4m
radius (r) of the cone = 1.2m
Density of the cone = 600kg/m³
Acceleration due to gravity, g = 9.8 m/s²
Now,
The total mass of the cone (m) = Density of the cone × volume of the cone
Volume of the cone = 
thus,
volume of the cone = 
= 6.03 m³
therefore, the mass of the cone = 600 Kg/m³ × 6.03 m³ = 3619.11 kg
The center of mass for the cone lies at the 
times the total height
thus,
center of mass lies at, h' = 
Now, the work gone (W) against gravity is given as:
W = mgh'
W = 3619.11kg × 9.8 m/s² × 1 = 35467.278 J
<span>C) mountains; two continental plates meet at a convergent boundary</span>
