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3 years ago

When a fluid is heated, what changes occur?

Physics

1 answer:

lisov135 [29]3 years ago

8 0

The answer to this question is A. It becomes less dense and begins to rise.

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Mercury is in the 80th position in the periodic table. How many protons does it have?

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Mercury has 80 protons.

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4 years ago

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A 50-g cube of ice, initially at 0.0°C, is dropped into 200 g of water in an 80-g aluminum container, both initially at 30°C.

MakcuM [25]

Answer:

b. 9.5°C

Explanation:

$m_i$ = Mass of ice = 50 g

$T_i$ = Initial temperature of water and Aluminum = 30°C

$L_f$ = Latent heat of fusion = $3.33\times 10^5\ J/kg^{\circ}C$

$m_w$ = Mass of water = 200 g

$c_w$ = Specific heat of water = 4186 J/kg⋅°C

$m_{Al}$ = Mass of Aluminum = 80 g

$c_{Al}$ = Specific heat of Aluminum = 900 J/kg⋅°C

The equation of the system's heat exchange is given by

$m_i(L_f+c_wT)+m_wc_w(T-T_i)+m_{Al}c_{Al}=0\\\Rightarrow 0.05\times (3.33\times 10^5+4186\times T)+0.2\times 4186(T-30)+0.08\times 900(T-30)=0\\\Rightarrow 1118.5T-10626=0\\\Rightarrow T=\dfrac{10626}{1118.5}\\\Rightarrow T=9.50022\ ^{\circ}C$

The final equilibrium temperature is 9.50022°C

4 0

3 years ago

What is average velocity?

Ber [7]

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time. According To a website on google

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3 years ago

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a plane travels 204 km, northeast in 15.0 minutes. It also increases elevation by 1.6 km, upward in the same amount of time. Wha

LUCKY_DIMON [66]

Answer:

230 m/s northeast, 1.8 m/s up

Explanation:

204 kilometres = 204000 metres

15.0 minutes = 900 seconds

Velocity = Distance / Time

= 204000 / 900

= 230 m/s northeast (to 2 sf.)

1.6km = 1600 metres

Velocity = 1600 / 900

= 1.8 m/s up (to 2 sf.)

Read more on Brainly.com - brainly.com/question/13863590#readmore

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3 years ago

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Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o

Harman [31]

Answer:

A) 1.55

B) 1.55

C) 12.92

D) 34.08

E) 57.82

Explanation:

The free body diagram attached, R is the radius of the wheel

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.

At the centre of the whee, torque due to B is given by

${\tau _2} = - {T_{\rm{B}}}R$

Similarly, torque due to A is given by

${\tau _1} = {T_{\rm{A}}}R$

The sum of torque at the pivot is given by

$\tau = {\tau _1} + {\tau _2}$

Replacing ${\tau _1}$ and ${\tau _2}$ by ${T_{\rm{A}}}R$ and $- {T_{\rm{B}}}R$ respectively yields

$\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}$

Substituting $I\alpha$ for $\tau$ in the equation $\tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

The angular acceleration of the wheel is given by $\alpha = \frac{a}{R}$

where a is the linear acceleration

Substituting $\frac{a}{R}$ for $\alpha$ into equation

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ we obtain

$\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

Net force on block A is

${F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}$

Net force on block B is

${F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g$

Where g is acceleration due to gravity

Substituting ${m_{\rm{B}}}a$ and ${m_{\rm{A}}}a$ for ${F_{\rm{B}}}$ and ${F_{\rm{A}}}$ respectively into equation $\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ and making a the subject we obtain

$\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}$

Since ${m_{\rm{B}}}$ = 3kg and ${m_{\rm{B}}}$ = 7kg

g=9.81 and R=0.12m, I=0.22 ${\rm{ kg}} \cdot {{\rm{m}}^2}$

Substituting these we obtain

$a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}$

$\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the linear acceleration of block A is 1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(B)

For block B

${a_{\rm{B}}} = {a_{\rm{A}}}$

Therefore, the acceleration of both blocks A and B are same

1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(C)

The angular acceleration is $\alpha = \frac{a}{R}$

$\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

(D)

Tension on left side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}$

(E)

Tension on right side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}$

6 0

3 years ago