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3 years ago

9

Assume that Michael's teacher has a rule that if a student talks out-of-turn three times in one day, that student must stay in f

or recess for the
rest of the week. Which of the following methods is she using to control student behavior?

a. folkway
b. sociobiology
c. informal sanction
d. formal sanction

1 answer:

olga_2 [115]3 years ago

4 0

D sorry if I get y’all wrong

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Suppose you have a piston in a cylinder where the gas temperature is 26.0° C and has a volume of 1.75 L. What is the volume of t

blsea [12.9K]

Answer:

1.62 L

Explanation:

Charle's law for ideal gases states that for a gas kept at constant pressure, the ratio between volume and temperature is constant:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where in this problem we have

$V_1 = 1.75 L$ is the initial volume of the gas

$T_1 = 26.0^{\circ}+273=299 K$ is the initial temperature of the gas

$V_2$ is the final volume of the gas

$T_2 = 3.0^{\circ}+273=276 K$ is the final temperature

Solving the equation for V2, we find

$V_2 = \frac{V_1}{T_1} T_2 = \frac{1.75 L}{299 K}(276 K)=1.62 L$

8 0

3 years ago

Read 2 more answers

A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

3 years ago

1. Encontrar la densidad de una sustancia, Si pesa 459 g y ocupa un volumen de 34.5 cmɜ.

ryzh [129]

Answer:

your language is to different to mine sorry

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3 years ago

Which type of element typically loses an electron to become an ion?

timofeeve [1]

Your answer should be metal

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3 years ago

How can you safely experiment with plants during a scientific investigation?

liubo4ka [24]

I would think the answer is c.

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3 years ago

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