What types of electromagnetic radiation does the sun emit?
1 answer:
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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>
</h2><h2>_____________________________________</h2><h2>DATA:</h2>
mass = m = 2kg
Distance = x = 6m
Force = 30N
TO FIND:
Work = W = ?
Velocity = V = ?
<h2>SOLUTION:</h2>According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.
To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,
Base is the x-axis of the graph which is Position i.e. 6m
Height is the y-axis of the graph which is Force i.e. 30N
So,
W = 90 J
The work done is 90 J.
According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.
Answer:
hello your question lacks some data and required diagram
G = 77 GPa, т all = 80 MPa
answer : required diameter = 252.65 * 10-^3 m
Explanation:
Given data :
force ( P ) = 660 -N force
displacement = 15 mm
G = 77 GPa
т all = 80 MPa
i) Determine the required diameter of shaft BC
considering the vertical displacement ( looking at handle DC from free body diagram )
D' = 0.3 sin∅ , where D = 0.015
hence ∅ = 2.8659°
calculate the torque acting at angle ∅ of CD on the shaft BC
Torque = 660 * 0.3 cos∅
= 660 * 0.3 * cos 2.8659 = 198 * -0.9622 = 190.5156 N
hello attached is the remaining part of the solution
