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2 years ago

7

A system absorbed 44 joules of heat from its surroundings. After doing work, the increase in the internal energy of the system i

s 32 joules. How much energy was used by the system to do useful work?

2 answers:

Lunna [17]2 years ago

6 0

Idk sorry maybe try @amyletbe

Mariulka [41]2 years ago

4 0

the answer is 12 joules on edg

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A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).

Tems11 [23]

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

$\sigma_{max}$ = Maximum stress

$\sigma_{min}$ = Minimum stress

$\sigma_m$ = Mean stress = 50 MPa

$\sigma_a$ = Stress amplitude = 225 MPa

Mean stress is given by

$\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}$

Stress amplitude is given by

$\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa$

$\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa$

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

$R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636$

The stress ratio is -0.63636

Stress range is given by

$\sigma_{max}-\sigma_{min}=450\ MPa$

Magnitude of the stress range is 450 MPa

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3 years ago

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Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900

Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron $m_{e}=9.11\times10^{-31}\ kg$

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}$

$E=0.582\ Mev$

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

$E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}$

$E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}$

$E=0.350\ Mev$

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

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