Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
ient of friction of 0.030 and a static coefficient of friction of 0.10.
2 answers:
Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
You might be interested in
Answer:
L/D= 112
Explanation:
Aerodynamics can be defined as the branch of dynamics which deals with the motion of air, their properties and the interaction between the air and solid bodies.
Aerodynamics law explains how an airplane is able to fly. There are four forces of flight, and they are; lift, weight, thrust and drag. The amount of lift generated by a wing divided by the aerodynamic drag is known as the lift to drag ratio.
Lift increases proportionally to the square of the speed.
The solutions to the question is the file attached to this explanation.
Lift,L= qC(l). S---------------------------(1).
and,
Drag,D = qC(d).S ----------------------(2).
Hence, Lift to drag ratio,L/D= C(l)/C(d).
Therefore, we have to compute various angle of attack.(check attached file)...
Then, (L/D) will then be equal to 112.
