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Oliga [24]
3 years ago
7

What is a crystalline solid?​

Physics
1 answer:
12345 [234]3 years ago
6 0

Answer:

A crystal or crystalline solid is a solid material whose constituents are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions

Explanation:

your welcome

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How much energy is stored in a 2.80-cm-diameter, 14.0-cm-long solenoid that has 200 turns of wire and carries a current of 0.800
ozzi

Answer:

The energy stored in the solenoid is 7.078 x 10⁻⁵ J

Explanation:

Given;

diameter of the solenoid, d = 2.80 cm

radius of the solenoid, r = d/2 = 1.4 cm

length of the solenoid, L = 14 cm = 0.14 m

number of turns, N = 200 turns

current in the solenoid, I = 0.8 A

The cross sectional area of the solenoid is given as;

A = \pi r^2\\\\A = \pi (0.014)^2\\\\A = 6.16*10^{-4} \ m^2

The inductance of the solenoid is given by;

L = \frac{\mu_0 N^2A}{l} \\\\L =  \frac{(4\pi*10^{-7})(200^2)(6.16*10^{-4})}{0.14}\\\\L = 2.212*10^{-4} \ H

The energy stored in the solenoid is given by;

E = ¹/₂LI²

E = ¹/₂(2.212 x 10⁻⁴)(0.8)²

E = 7.078 x 10⁻⁵ J

Therefore, the energy stored in the solenoid is 7.078 x 10⁻⁵ J

8 0
3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

8 0
3 years ago
A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.
Levart [38]

Answer:

I don't exactly know what you learned but it could be because of more friction or the bus was running out of gas.

4 0
3 years ago
Which question cannot be answered through making measurements?
Bond [772]
A. is the answer for this question
5 0
3 years ago
In this problem, you will practice applying this formula to several situations involving angular acceleration. In all of these s
riadik2000 [5.3K]

Answer:

Part a)

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

Explanation:

As we know that the see saw bar is massless so here torque due to two masses is given as

\tau = I\alpha

here we will have

\tau = (m_1g - m_2g)(\frac{L}{2})

now we will have inertia of two masses given as

I = (m_1 + m_2)(\frac{L}{2})^2

now we have

I = (m_1 + m_2)\frac{L^2}{4}

now the angular acceleration is given as

\alpha = \frac{\tau}{I}

so we have

\alpha = \frac{2(m_1 - m_2)g}{(m_1 + m_2)L}

Part b)

Now if the rod is not massles then we will have total inertia given as

I = (m_1 + m_2)(\frac{L}{2})^2 + \frac{m_{bar}L^2}{12}

so we will have

I = (m_1 + m_2)\frac{L^2}{4} + \frac{m_{bar}L^2}{12}

now the acceleration is given as

\alpha = \frac{\tau}{I}

\alpha = \frac{6(m1 - m_2)g}{3(m_1 + m_2)L + m_{bar}L}

7 0
3 years ago
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