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2 years ago

Why is the battery run flat with reference to the energy transformations that take place

Physics

1 answer:

andreev551 [17]2 years ago

7 0

Answer:

Rechargeable batteries can still go flat after repeated use because the materials involved in the reaction lose their ability to charge and regenerate

Explanation:

they go flat because they a rechargeable and sometimes don't always work

thank you hope u have a good day

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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

What is the volume of an object that has Mass=100g and Density= 745g/mL.

lana66690 [7]

Answer:0.13ml

Explanation:

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3 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

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3 years ago

Which of the following is an example of chemical change?

Maru [420]

Answer:

Explanation:

all the others r physical

8 0

3 years ago

Read 2 more answers

3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ

WITCHER [35]

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

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3 years ago

Why is the battery run flat with reference to the energy transformations that take place​

Why is the battery run flat with reference to the energy transformations that take place