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Sindrei [870]
3 years ago
7

Infer whether a system can have kinetic energy and potential energy at the same time

Physics
1 answer:
trapecia [35]3 years ago
7 0
A system can have kinetic and potential energy, for example, an object falling but it hasn't reached the ground. It's still in the air, and has "potential" to go faster or hit something on it's way down, but since it's moving it also has kinetic energy. 
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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

8 0
3 years ago
Free electrons may be transferred between bodies by:
Tasya [4]
<span>Free electrons may be transferred between bodies by "Contact Action"

When positively charged body comes close to a negatively charged body, then electron flows from positively to negatively charged body

In short, Your Answer would be Option B

Hope this helps!</span>
8 0
3 years ago
Read 2 more answers
R=1m.<br> Vt=+- 8m/s<br> atot (tan<br> √3)
erica [24]

What What What What...... error

7 0
2 years ago
This is the phase of matter with no fixed shape but fixed volume.
Nataly_w [17]
A liquid is a matter that neither has a fixed shape but fixed volume...in college really need to know this.

GOOD LUCK
6 0
3 years ago
Read 2 more answers
A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re
lbvjy [14]

Answer:

The beam of light is moving at the peed of:

\frac{dy}{dt} = \frac{80\pi}{3} km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, \omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi    (1)

Now, in the right angle in the given fig.:

tan\theta' = \frac{y}{3}

Now, differentiating both the sides w.r.t t:

\frac{dtan\theta'}{dt} = \frac{dy}{3dt}

Applying chain rule:

\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}

sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}

Now, using tan\theta = \frac{1}{m} and y = 1 in the above eqn, we get:

(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}

Also, using eqn (1),

8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}

\frac{dy}{dt} = \frac{80\pi}{3}

7 0
3 years ago
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