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2 years ago

Why does it take longer to boil a kettle of water than to warm the same kettle of water to a lower temperature?

Physics

1 answer:

lys-0071 [83]2 years ago

6 0

Answer:

The reason that it takes longer to get the water to boiling temperature than it is to cool it down again is because heating in the most simple sense is inefficient and will cause a lot if energy lost while cooling is to be turn's into quite a efficient process.

Explanation:

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What is a stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second?

julia-pushkina [17]

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

Discharge rate = velocity * area

= velocity * depth * width

= 2 * 2 * 10 = 40 $m^{3}$ /sec

A stream’s discharge rate if it has a width of 10 meters, a depth of 2 meters, and a velocity of 2 meters per second will be 40 $m^{3}$ /sec .

learn more about discharge rate

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2 years ago

A 90. 0-kg ice hockey player hits a 0. 150-kg puck, giving the puck a velocity of 45. 0 m/s. If both are initially at rest and i

Mice21 [21]

The distance traveled by the hockey player is 0.025 m.

<h3>The principle of conservation of linear momentum;</h3>

The principle of conservation of linear momentum states that, the total momentum of an isolated system is always conserved.

The final velocity of the hockey play is calculated by applying the principle of conservation of linear momentum;

$m_1v_1 = m_2 v_2\\\\v_1 = \frac{m_2 v_2}{m_1} \\\\v_1 = \frac{0.150 \times 45}{90} \\\\v_1 = 0.075 \ m/s$

The time taken for the puck to reach 15 m is calculated as follows;

$t = \frac{d}{v} \\\\t = \frac{15\ m}{45 \ m/s} \\\\t = 0.33 \ s$

The distance traveled by the hockey player at the calculated time is;

$d = vt\\\\d = 0.075 \ m/s \ \times 0.33 \ s\\\\d = 0.025 \ m$

Learn more about conservation of linear momentum here: brainly.com/question/7538238

4 0

2 years ago

You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential

Dahasolnce [82]

True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...

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2 years ago

Atmospheric pressure decreases as altitude increases. in other words, there is more air pushing down on you at sea level, and th

Nutka1998 [239]

At sea level, the size amid the 2 alkanes lets for pentane to simmer at a lower temperature than hexane. Phenol has a higher boiling point due to hydrogen bonding High altitude would have the same order while low pressure only cuts the temperature at which a solvent boils. Boiling has to do with molecular size, the occurrence/nonappearance of hydrogen bonds, and other steric issues.
So the answer would be pentane high altitude, hexane high altitude, hexane sea level, hexanol sea level. In order of boil first to boil last. This is clarified because altitude has a better effect on vapor pressure (and hence boiling points) than inter-molecular forces.

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3 years ago

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