Kinetic energy: the energy of motion
Work: the change in kinetic energy
Power: the rate of work done
Explanation:
The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:
where
m is the mass of the object
v is its speed
The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:
where
is the final kinetic energy
is the initial kinetic energy
Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as
where
W is the work done
t is the time elapsed
Learn more about kinetic energy, work and power:
brainly.com/question/6536722
brainly.com/question/6763771
brainly.com/question/6443626
brainly.com/question/7956557
#LearnwithBrainly
Answer:
τ = 132.773 lb/in² = 132.773 psi
Explanation:
b = 12 in
F = 60 lb
D = 3.90 in (outer diameter) ⇒ R = D/2 = 3.90 in/2 = 1.95 in
d = 3.65 in (inner diameter) ⇒ r = d/2 = 3.65 in/2 = 1.825 in
We can see the pic shown in order to understand the question.
Then we get
Mt = b*F*Sin 30°
⇒ Mt = 12 in*60 lb*(0.5) = 360 lb-in
Now we find ωt as follows
ωt = π*(R⁴ - r⁴)/(2R)
⇒ ωt = π*((1.95 in)⁴ - (1.825 in)⁴)/(2*1.95 in)
⇒ ωt = 2.7114 in³
then the principal stresses in the pipe at point A is
τ = Mt/ωt ⇒ τ = (360 lb-in)/(2.7114 in³)
⇒ τ = 132.773 lb/in² = 132.773 psi
The total work done is 5980 Joules and the power expended is 57 Watts.
<h3>What is work done?</h3>
The work done is the work done in the gravitational field as the bucket is raised up Thus work required to remove the bucket Wb;
Wb = 13.9 kg * 25.9 m * 9.8 m/s^2 = 3530 Joules
Height of the center of mass of chain = 25.9 / 2 = 12.95 m
Work done by the chain Wc;
Wc = 12.95 * 19.3 * 9.8 = 2450 Joules
Total work = 3530 + 2450 = 5980 Joules
Power expended = W / t = 5980 J / 105 sec = 57 J/s = 57 Watts
Learn more about work done:brainly.com/question/13662169
#SPJ1
Answer:
A block device is a computer data storage device that supports reading and (optionally) writing data in fixed-size blocks, sectors, or clusters. These blocks are generally 512 bytes or a multiple thereof in size
For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
