Answer:
The force of static friction acting on the luggage is, Fₓ = 180.32 N
Explanation:
Given data,
The mass of the luggage, m = 23 kg
You pulled the luggage with a force of, F = 77 N
The coefficient of static friction of luggage and floor, μₓ = 0.8
The formula for static frictional force is,
Fₓ = μₓ · η
Where,
η - normal force acting on the luggage 'mg'
Substituting the values in the above equation,
Fₓ = 0.8 x 23 x 9.8
= 180.32 N
Hence, the minimum force require to pull the luggage is, Fₓ = 180.32 N
Answer:
b) true. The jobs are equal
Explanation:
The work on a body is the scalar product of the force applied by the distance traveled.
W = F. d
Work is a scalar, the work equation can be developed
W = F d cos θ
Where θ is the angle between force and displacement
Let's apply these conditions to the exercise
a) False, if we see the expression d cosT is the projection of the displacement in the direction of the force, so there may be several displacement, but its projection is always the same
b) true. The jobs are equal dx = d cosθ
c) False, because the force is equal and the projection of displacement is the same
d) False, knowledge of T is not necessary because the projection of displacement is always the same
e) False mass is not in the definition of work
Answer:
Part a)
Part b)
T = 4.68 s
Explanation:
Part a)
Shell is fired at speed of 40 m/s at angle of 35 degree
so here we have
since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero
so at the highest point the speed is given
Part b)
After completing the motion we know that the displacement of the object will be zero in Y direction
so we have
In one quadrant there are 90 degrees.
1. Mitosis and Meiosis are both ways cells duplicate DNA
2. They both go through PMAT ( prophase, metaphase, anaphase and telophase)
