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Nady [450]
3 years ago
12

If there is "waste" energy, does the Law of Conservation of Energy still apply? please don't type something random if so i'll ju

st report it.​
Physics
1 answer:
FinnZ [79.3K]3 years ago
8 0

Explanation:

Yes, the law of conservation of energy still applies even if there is waste energy.

The waste energy are the transformation products of energy from one form to another.

According to the law of conservation of energy "energy is neither created nor destroyed by transformed from one form to another in a system".

But of then times, energy is lost as heat or sound within a system.

  • If we take into account these waste energy, we can see that energy is indeed conserved.
  • The sum total of the energy generated and those produced will be the same if we factor in other forms in which the energy has been transformed into.
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A rod has a radius of 10 mm is subjected to an axial load of 15 N such that the axial strain in the rod is ????௫ = 2.75*10-6, de
EleoNora [17]

Answer:

Knowing we only have one load applied in just one direction we have to use the Hooke's law for one dimension

ex = бx/E

бx = Fx/A = Fx/πr^{2}

Using both equation and solving for the modulus of elasticity E

E = бx/ex = Fx / πr^{2}ex

E = \frac{15}{pi (10 * 10^{-3})^{2} * 2.75 * 10^{-6}    } = 17.368 * 10^{9} Pa = 17.4 GPa

Apply the Hooke's law for either y or z direction (circle will change in every direction) we can find the change in radius

ey = \frac{1}{E} (бy - v (бx + бz)) = -\frac{v}{E}бx

= \frac{vFx}{Epir^{2} } = \frac{0.23 * 15}{pi (10 * 10^{-3)^{2} } * 17.362 * 10^{9}  } = -0.63 *10^{-6}

Finally

ey = Δr / r

Δr = ey * r = 10 * -0.63* 10^{-6} mm = -6.3 * 10^{-6} mm

Δd = 2Δr = -12.6 * 10^{-6} mm

Explanation:

5 0
4 years ago
Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the
Daniel [21]

Answer: 4.8 s

Explanation:

We have the following data:

m=180 kg the mass of the raft

F=75 N the force applied by Sawyer

V=2 m/s the raft's final speed

V_{o}=0 m/s the raft's initial speed (assuming it starts from rest)

We have to find the time t

Well, according to Newton's second law of motion we have:

F=m.a (1)

Where a is the acceleration, which can be expressed as:

a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}} (2)

Substituting (2) in (1):

F=m\frac{V-V_{o}}{t-t_{o}} (3)

Where t_{o}=0

Isolating t from (3):

t=\frac{m(V-V_{o})}{F} (4)

t=\frac{180 kg(2 m/s-0 m/s)}{75 N}

Finally:

t=4.8 s

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3 years ago
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salantis [7]

\boxed{\sf \tau=rFsin\theta}

Put values

\\ \rm\hookrightarrow 40=2Fsin40

\\ \rm\hookrightarrow Fsin40=20

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