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3 years ago

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Which states of matter are significantly compressible?

1 answer:

andreev551 [17]3 years ago

5 0

Hello.

The gaseous state is the more compressible state, because it has the volume of its container.

The liquid state is virtually incompressible, and the solid state compression is very small.

The plasma is another state that has high compression, but in this case the matter is not bound(we don't have the proton in the core of the atom)

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A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki

Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

$\tau=I*\alpha$

given

$\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?$

$\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2$

the angular velocity is

$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

$KE= 1/2*I* \omega^2$

$KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J$

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3 years ago

an electron is moving with a speed of 0.85c in a direction opposite to that of photon. calculate the relative velocity of the el

ololo11 [35]

Answer:

1.85c

Explanation:

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2 years ago

A ball is launched with an initial horizontal velocity of 10.0 meters per second. It takes 500 milliseconds for the ball to reac

hichkok12 [17]

Answer:

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For initial vertical velocity just use u sin theta.

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3 years ago

Haley's swim team hosted a home meet and needed volunteers. How did she reveal to her coach that she is not a team player?

vladimir2022 [97]

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3 years ago

Read 2 more answers

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical

alexdok [17]

Answer:

The critical stress required for the propagation of an initial crack $\sigma_{c}$ = 21.84 M pa

Explanation:

Given data

Modulus of elasticity E = 225 × $10^{9}$ $\frac{N}{m^{2} }$

Specific surface energy for magnesium oxide is $\gamma_{s}$ = 1 $\frac{J}{m^{2} }$

Crack length (a) = 0.3 mm = 0.0003 m

Critical stress is given by $\sigma_{c}^{2} }$ = $\frac{2 E \gamma}{\pi a}$ -------- (1)

⇒ 2 E $\gamma_{s}$ = 2 × 225 × $10^{9}$ × 1 = 450 × $10^{9}$

⇒ $\pi$ a = 3.14 × 0.0003 = 0.000942

⇒ Put these values in equation 1 we get

⇒ $\sigma_{c}^{2} }$ = $\frac{450 }{0.000942} 10^{9}$

⇒ $\sigma_{c}^{2} }$ = 4.77 × $10^{14}$

⇒ $\sigma_{c}$ = 2.184 × $10^{7}$ $\frac{N}{m^{2} }$

⇒ $\sigma_{c}$ = 21.84 $\frac{N}{mm^{2} }$

⇒ $\sigma_{c}$ = 21.84 M pa

This is the critical stress required for the propagation of an initial crack.

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3 years ago