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Murljashka [212]

3 years ago

11

A jeweler working with a heated 47 g gold ring must lower the ring's temperature to make it safe to handle. If the ring is initi

ally at 99ºC, what mass of water at 25ºC is needed to lower the ring's temperature to 38ºC?
5.9 g
3.7 g
2.2 g
6.8 g

2 answers:

Gelneren [198K]3 years ago

6 0

Mass of gold m₁ = 47 g

Initial temperature of gold T₁ = 99 C

Specific heat of gold C₁ = 0.129 J/gC

final temperature T₂ = 38 C

Heat needed by the gold to cool down

Q =m₁ * C₁* ( T₁ - T₂)

Q = (47)(0.129)(99-38)

Q = 369.843 J

This heat will be given by the water

we need to find out mass of water m₂

and initial temperature of water is T₃ = 25 C

Specific heat of water C₂ = 4.184 J/gC

Q = m₂*C₂*(T₂ - T₃)

369.843 = m₂(4.184)(38-25)

m₂ = 6.8 g

harkovskaia [24]3 years ago

3 0

The answer is D, 6.8 g.

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

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Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

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$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

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car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

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Answer:

The force of car 3 on car 2 ≈ 1810.82 N

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150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

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$v_f$ = 0.1627 m/s

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Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

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An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

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We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

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Now we can use the second equation to solve for the distance travelled by the airplane:

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3 years ago