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dusya [7]
3 years ago
7

A density of solid 13g/cm³ at a temperature of 25°c was now heated to a temperature of 150°c find the new density of the solid ?

Given linear expansivity to be 2 × 10⁻⁵ ?​
Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

13.4225 g/cm³

Explanation:

Given that:

Linear expansivity α = 2 * 10^-5

Change in temperature dt = t2 - t1 = (150 - 25)°c = 125°c

Initial density (d1) = 13

New density d2 =?

Using the relation :

α = (d2 - d1) / d1 * dt

d2 - d1 = d1 * dt * α

d2 = d1 + d1*dt*α

d2 = d1(1 + d1*dt*α)

d2 = 13( 1 + (13*125*2*10^-5))

d2 = 13(1 + 0.0325)

d2 = 13(1.0325)

d2 = 13.4225

d2 = 13.4225 g/cm³

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1)

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Answer:

d. 12.25J

------------------------

2)

According to the conservation of energy:

\begin{gathered} E1=E2 \\ so: \\ E2=12.25J \end{gathered}

Answer:

b. 12.25

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3)

P1=m_1v_2+m_2v_2=+1.75-1.75=0

Answer:

d. 0 kg∙m/s

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4)

Using conservation of momentum:

\begin{gathered} P1=P2 \\ so: \\ P2=0 \end{gathered}

Answer:

b. 0 kg•m/s

3 0
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Answer:

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\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

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6 0
3 years ago
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Answer:

0

Explanation:

Since no distance is given, the force is not doing any work

No work is done by the man since we do not know the distance or displacement.

Work is only said to be done when the force applied on an object moves it through a particular distance.

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Since no distance is given in this problem, we can as well assume that the force applied is doing no work on the object.

4 0
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