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Y_Kistochka [10]

3 years ago

10

Under what conditions is energy conserved

1 answer:

Alinara [238K]3 years ago

8 0

Answer:

If only internal forces are doing work (no work done by external forces), then there is no change in the total amount of mechanical energy. The total mechanical energy is said to be conserved. ... In these situations, the sum of the kinetic and potential energy is everywhere the same.

Explanation:

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When a cyclist moves downhill without pedalling, what type of energy does he gain?

Vesna [10]

Answer:

He is gaining kinetic energy and losing potential energy

Explanation:

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2 years ago

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The ideal gas law includes all of the following properties, except the

Verizon [17]

The correct answer among the choices is option D. Density is not one of the properties included in the ideal gas law. The law is expressed as: PV=nRT. As we can see, the pressure, the volume and the temperature of the gas are included in the law.

3 0

3 years ago

The graphic organizer above shows that the properties of waves are influenced by the energy of waves. Name 2 properties of waves

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Amplitude: the height of the wave<span>, measured in meters
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3 0

2 years ago

A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with

Julli [10]

Explanation:

Given that,

Capacitor $C=1.66\ \mu F$

Resistor $R=80.0\ \Omega$

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

$f_{c}=\dfrac{1}{2\pi R C}$

Where, R = resistor

C = capacitor

Put the value into the formula

$f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}$

$f_{c}=1198.45\ Hz$

(B). We need to calculate the $V_{R}$ when $f = \dfrac{1}{2f_{c}}$

Using formula of $V_{R}$

$V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})$

Put the value into the formula

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=2.280\ Volt$

(C). We need to calculate the $V_{R}$ when $f = f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=3.606\ Volt$

(D). We need to calculate the $V_{R}$ when $f = 2f_{c}$

Using formula of $V_{R}$

$V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})$

$V_{R}=4.561\ Volt$

Hence, This is the required solution.

8 0

3 years ago

20.0 moles, 1840 g, of a nonvolatile solute, C 3H 8O 3 is added to a flask with an unknown amount of water and stirred. The solu

Anastasy [175]

Answer:

0.144 kg of water

Explanation:

From Raoult's law,

Mole fraction of solvent = vapor pressure of solution ÷ vapor pressure of solvent = 423 mmHg ÷ 528.8 mmHg = 0.8

Let the moles of solvent (water) be y

Moles of solute (C3H8O3) = 2 mole

Total moles of solution = moles of solvent + moles of solute = (y + 2) mol

Mole fraction of solvent = moles of solvent/total moles of solution

0.8 = y/(y + 2)

y = 0.8(y + 2)

y = 0.8y + 1.6

y - 0.8y = 1.6

0.2y = 1.6

y = 1.6/0.2 = 8

Moles of solvent (water) = 8 mol

Mass of water = moles of water × MW = 8 mol × 18 g/mol = 144 g = 144/1000 = 0.144 kg

7 0

2 years ago