static electricity and friction
The work done by the applied force on the block against the frictional force is 15.75 J.
<h3>
Work done by the applied force</h3>
The work done by the applied force is calculated as follows;
W = Fd
F - Ff = ma
where;
- F is applied force
- Ff is frictional force
Fcos(37) - μmgsin(37) = ma
Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)
0.799F - 7.077 = 0.8
F = 9.86 N
W = Fdcosθ
W = 9.86 x 2 x cos(37)
W = 15.75 J
Thus, the work done by the applied force on the block against the frictional force is 15.75 J.
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<span> Displacement of the medium perpendicular to the direction of propagation of the wave. that would be your answer</span>
The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.
The given parameters:
- <em>Throat area of the nozzle, </em>
<em> = 10 cm² = 0.001 m²</em> - <em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
- <em>Air pressure at sea level = 101.325 kPa</em>
The ratio of the areas of the converging-diverging nozzle is calculated as follows;

From supersonic isentropic table, at
, we can determine the following;

The tank pressure is calculated as follows;

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.
Learn more about converging-diverging nozzle design here: brainly.com/question/13889483
Answer:
The crate's coefficient of kinetic friction on the floor is 0.23.
Explanation:
Given that,
Mass of the crate, m = 300 kg
One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 320 N using a rope connected to the crate.
The crate slides with a constant speed. It means that the net force acting on it is 0. Net force acting on it is given by :

So, the crate's coefficient of kinetic friction on the floor is 0.23.