Here when an object is placed on the level floor then in that case there are two forces on the object
1). Weight of object downwards (mg)
2). Normal force due to floor which will counterbalance the weight (N)
so when no force is applied on the box at that time normal force is counter balanced by weight.
Now here it is given that A person tried to lift the box upwards
So now there are two forces on the box
1). Applied force of person
2). Normal force due to ground
So now these two forces will counter balance the weight of the crate
So we can write an equation for force balance like
![F_g = F_n + F_a](https://tex.z-dn.net/?f=F_g%20%3D%20F_n%20%2B%20F_a)
given that
![F_g = mg](https://tex.z-dn.net/?f=F_g%20%3D%20mg)
here
m = 30 kg and
g = acceleration due to gravity = 10 m/s^2
![F_n = 150 N](https://tex.z-dn.net/?f=F_n%20%3D%20150%20N)
now from above equation
![30*10 = 150 + F_a](https://tex.z-dn.net/?f=30%2A10%20%3D%20150%20%2B%20F_a)
![F_a = 300 - 150 = 150 N](https://tex.z-dn.net/?f=F_a%20%3D%20300%20-%20150%20%3D%20150%20N)
So force applied by the person must be 150 N
the answer is CaO because that's what my homework says is correct
Answer:
0.466 (3 sig. fig.)
Explanation:
Frictional force acting on the box = 5.00×10^2xsin25
Normal force acting on the box = 5.00×10^2xcos25
coefficient of friction = 0.466 (3 sig. fig.)
Check-call-Care if you are training to be a life guard.
B would be an example of vaporization (liquid to gas).
———————
A is an example of deposition (gas to solid); C is an example of condensation (gas to liquid); and D is an example of condensation, deposition, or freezing—depending on the type of cloud.