For rotational equilibrium of the door we can say that torque due to weight of the door must be counter balanced by the torque of external force

here weight will act at mid point of door so its distance is half of the total distance where force is applied
here we know that

now we will have


so our applied force is 72.5 N
0.4823 m/s
The initial velocity u1 of the ball=0
From the law of conservation of linear momentum.
m1u1+m2u2=m1v1+m2v2
(160×0)+(170×u1)=(160×0.3)+(170×0.2)
u1=0.4823m/s
Answer:
<em>2.78m/s²</em>
Explanation:
Complete question:
<em>A box is placed on a 30° frictionless incline. What is the acceleration of the box as it slides down the incline when the co-efficient of friction is 0.25?</em>
According to Newton's second law of motion:

Where:
is the coefficient of friction
g is the acceleration due to gravity
Fm is the moving force acting on the body
Ff is the frictional force
m is the mass of the box
a is the acceleration'
Given

Required
acceleration of the box
Substitute the given parameters into the resulting expression above:
Recall that:

9.8sin30 - 0.25(9.8)cos30 = ax
9.8(0.5) - 0.25(9.8)(0.866) = ax
4.9 - 2.1217 = ax
ax = 2.78m/s²
<em>Hence the acceleration of the box as it slides down the incline is 2.78m/s²</em>
Answer:
Spring constant, k = 0.3 N/m
Explanation:
It is given that,
Force acting on DNA molecule, 
The molecule got stretched by 5 nm, 
Let k is the spring constant of that DNA molecule. It can be calculated using the Hooke's law. It says that the force acting on the spring is directly proportional to the distance as :



k = 0.3 N/m
So, the spring constant of the DNA molecule is 0.3 N/m. Hence, this is the required solution.