Ask question

Ask question

All categories

Alenkasestr [34]

3 years ago

14

Christopher conducts an experiment in which he tests how much sugar dissolves at different temperatures of water. One step requi

res him to use water with a temperature of 32.5°C. What is this temperature in kelvins? Record your answer to the nearest tenth. K

2 answers:

sveticcg [70]3 years ago

5 0

305.5 i just took the quiz

disa [49]3 years ago

4 0

Explanation:

When we have to convert temperature from degree celsius to kelvin then we add 273 into the given degree celsius temperature.

Therefore, convert $32.5^{o}C$ into kelvin as follows.

Temperature in kelvin = Temperature in degree celsius + 273

= (32.5 + 273) K

= 305.5 K

Thus, we can conclude that temperature in kelvins is 305.5 K.

You might be interested in

Based on the information in the graph, which of the atoms listed below is the most stable?

elixir [45]

Based on the information in the graph, the atom which is listed below is the most stable would be A. Oxygen-16 (O-16).

8 0

3 years ago

Read 2 more answers

A current-carrying loop of wire lies flat on a horizontal tabletop. When viewed from above, the current moves around the loop in

Oksana_A [137]

Answer: i dont do physics yet lol

Explanation:

6 0

2 years ago

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello

klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0

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

2 years ago

Could someone answer this?

Nadusha1986 [10]

Answer:

Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.

Now,

The maximum power of each resistance is 16 W

The 4Ω resistor is linked in series with the circuit.

so, P o w e r = I

two

R, here i is the current through the resistor resistor R

1 6 = I

two

∗ 4 Ω

i = 2A

Now 2A passes through parallel resistors of 8Ω resistance.

we know that, in parallel, the potential difference must be constant,

the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is

2 A

two

.

= 1 A

So finally the current through the 4Ω resistor = 2 A

current through each 8Ω resistor = 1 A

Explanation:

I hope this answer has helped you

6 0

3 years ago