Since rope is parallel to the inclined plane so here we can say that net force parallel to the person which is pulling upwards must counterbalance the component of weight of the person.
Now here we will do the components of the weight of the person
given that weight of the person = 500 N
now its components are


now here as we can say that one of the component is balanced here by the normal force perpendicular to plane
while the other component of the weight is balanced by the force applied on the rope
So here the force applied on the rope will be given as


so it apply 300 N force along the inclined plane
Answer:The mass of an object is 52 kg.
Explanation:
Gravitational force on the object ,F=510 N
Acceleration due to gravity = g = 
Mass of the object = m
Force = mass × acceleration


The mass of an object is 52 kg.
<span>1 cal = 4,185 J
1 kcal = 1*10^3 cal
or
=1000 cal</span>
Answer:
L=31.9 mm
δ = 0.22 mm
Explanation:
Given that
v= 14 m/s
ρ=997 kg/m³
μ= 0.891 × 10⁻3 kg/m·s
As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.



L=0.0319 m
L=31.9 mm
The thickness of the boundary layer at that location L given as


δ = 0.00022 m
δ = 0.22 mm