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3 years ago

12

As a cat pounces on a mouse, its muscles consume 10 units of potential energy (which the cat previously gained from the food it

consumed). However, the pounce itself only required 4 units of kinetic energy. How many units of energy were dissipated as heat?

1 answer:

taurus [48]3 years ago

7 0

6 units of energy were transferred into heat.

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1. 412.9 g of dry ice sublimes at room temperature. a. What’s changing? --- sublimation b. What constant will you use? ----- 25.

hoa [83]

1. 236 kJ

a. The phase (or state of matter) of the substance: from solid state to gas state (sublimation)

b. The enthalphy of sublimation, given by: $\lambda=571 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of dry ice and $\lambda$ is the enthalpy of sublimation

d. The energy is being absorbed, because the heat is transferred from the environment to the dry ice: as a consequence, the bonds between the molecules of dry ice break and then move faster and faster, and so the substance turns from solid into gas directly.

e. The amount of energy being transferred is

$Q=m\lambda=(412.9 g)(571 J/g)=2.36\cdot 10^5 J=236 kJ$

2. 165 kJ

a. The phase (or state of matter) of the substance: from gas state to liquid state (condensation)

b. The latent heat of vaporisation of water, given by $\lambda=2260 J/g$

c. The equation to use will be $Q=m\lambda$ , where m is the mass of steam that condenses and $\lambda$ is the latent heat of vaporisation

d. The energy is being released, since the substance turns from a gas state (where molecules move faster) into liquid state (where molecules move slower), so the internal energy of the substance has decreased, therefore heat has been released

e. The amount of energy being transferred is

$Q=m\lambda=(72.9 g)(2260 J/g)=1.65\cdot 10^5 J=165 kJ$

3. 3.64 kJ

a. Only the temperature of the substance (which is increasing)

b. The specific heat capacity of silver, which is $C_s = 0.240 J/gC$

c. The equation to use will be $Q=m C_s \Delta T$ , where m is the mass of silver, Cs is the specific heat capacity and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the silver, since its temperature increases, this means that its molecules move faster so energy should be provided to the silver by the surroundings

e. The amount of energy being transferred is

$Q=m C_s \Delta T=(39.2 g)(0.240 J/gC)(412.9^{\circ}C-25.9^{\circ}C)=3641=3.64 kJ$

4. 89 kJ

a. Both the phase of the substance (from solid to liquid) and then the temperature

b. The latent heat of fusion of ice: $\lambda=334 J/g$ and the specific heat capacity of water: $C_s=4.186 J/gC$

c. The equation to use will be $Q=m\lambda + m C_s \Delta T$ , where m is the mass of ice, $\lambda$ the latent heat of fusion of ice, Cs is the specific heat capacity of water and $\Delta T$ the increase in temperature

d. The energy is being absorbed by the ice, at first to break the bonds between the molecules of ice and to cause the melting of ice, and then to increase the temperature of the water

e. The amount of energy being transferred is

$Q=m\lambda +m C_s \Delta T=(156.3 g)(334 J/g)+(156.3 g)(4.186 J/gC)(56.232^{\circ}C-0^{\circ}C)=8.9\cdot 10^4 J=89 kJ$

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3 years ago

A Venturi tube may be used as a fluid flowmeter. Suppose the device is used at a service station to measure the flow rate of gas

leonid [27]

Answer

given,

flow rate = p = 660 kg/m³

outer radius = 2.8 cm

P₁ - P₂ = 1.20 k Pa

inlet radius = 1.40 cm

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₁² v₂

$v_1= \dfrac{r_1^2}{r_2^2} v_2$

$v_1= \dfrac{1.4^2}{2.8^2} v_2$

$v_1= 0.25 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$\Delta P = \dfrac{1}{2}\rho (v_2^2-(0.25 v_2)^2)$

$\Delta P = \dfrac{1}{2}\rho v_2^2 (1 - 0.0625)$

$v_2=\sqrt{\dfrac{2\Delta P}{\rho(1 - 0.0625)}}$

$v_2=\sqrt{\dfrac{2\times 1200}{660 \times(1 - 0.0625)}}$

v₂ = 1.97 m/s

b) fluid flow rate

Q = A₂ V₂

Q = π (0.014)² x 1.97

Q = 1.21 x 10⁻³ m³/s

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4 years ago

Batteries are rated in terms of ampere-hours (A·h). For example, a battery that can produce a current of 2.00 A for 3.00 h is ra

Reptile [31]

(a) 423 J

The power of the battery is the ratio between the total energy stored (E) and the time elapsed (t):

$P=\frac{E}{t}$

However, the power is also the product of the voltage (V) and the current (I):

$P=VI$

Linking the two equations together,

$\frac{E}{t}=VI\\E=VIt$

Since we know:

V = 9.0 V

$I \cdot t = 47.0 A\cdot h$

We can calculate the total energy:

$E=(9.0 V)(47 A \cdot h)=423 J$

(b) $7.79\cdot 10^{-6}$ dollars

The battery has a total energy of E = 423 J. (2)

1 Watt (W) is equal to 1 Joule (J) per second (s):

$1 W = \frac{1 J}{1 s}$

so 1 kW corresponds to 1000 J/s:

$1 kW = \frac{1000 J}{1 s}$

Multiplying both side by 1 hour (1 h):

$1 kW \cdot h = \frac{1000 J}{1 s} 1 h$

and $1 h = 3600 s$ , so

$1 kWh = \frac{1000 J}{1 s}\cdot 3600 s =3.6\cdot 10^6 J$

So we find the conversion between kWh and Joules. So now we can convert the energy from Joules (2) into kWh:

$1 kWh = 3.6\cdot 10^6 J = x : 423 J\\x=\frac{1 kWh \cdot 423 J}{3.6\cdot 10^6 J}=1.18\cdot 10^{-4}kWh$

And since the cost is $0.0660 per kilowatt-hour, the total cost will be

$C=$0.0660\cdot 1.18\cdot 10^{-4} kWh=7.79\cdot 10^{-6}$ dollars

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4 years ago

A form letter can be customized by using different fields in a __________.

Juli2301 [7.4K]

The answer is C. Email

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3 years ago

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A solid conducting sphere is placed in an external uniform electric field. With regard to the electric field on the sphere's int

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There is no electric field on the interior of the conducting sphere..

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