All categories

3 years ago

How would the plane strain fracture toughness of a metal be expected to change with rising temperature?

Physics

1 answer:

aksik [14]3 years ago

4 0

Answer:

Increase

Explanation:

The plane strain fracture toughness of a metal is expected to increase with rising temperature.

You might be interested in

Why is the rotation system used in volleyball? Physics was the closest thing to PE

tankabanditka [31]

Answer:

A rotation occurs after every side out, which is when the receiving team gains the right to serve by winning a rally. ... The new serving team will rotate clockwise one spot. The purpose of this is to rotate all the players through the serving position. If you continue winning points, you stay in position.

3 0

3 years ago

Read 2 more answers

HELP ASAP!!!! 20 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!

Alexxandr [17]

Answer:

I would have to say the answer is D

Explanation:

because the angle is being changed using the ray box.

8 0

3 years ago

What is the speed of rocket that travels 9km in 10 seconds

zysi [14]

Answer:

900

Explanation:

v = s / t = 9000m / 10 s = 900m/s

4 0

3 years ago

Read 2 more answers

PPPLLLLSSS HELP ME Identify the following as visible light, radio, ultraviolet, gamma, micro, infrared waves this is for the sec

Lorico [155]

I’m trying to get things expanded graph explanation sorry

6 0

2 years ago

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago