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3 years ago

An electric circuit consists of light bulb with resistance of 10.0 ohm is connected to a

Physics

1 answer:

Usimov [2.4K]3 years ago

8 0

Answer:

P=I²R

Explanation:

P=2*10^-3 square times 10

4 times ten raise to power minus 5

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The state of matter in which the molecules are closest together is the _____.

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Solid phase. The atoms are tightly packed and vibrate.

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4 years ago

The rotational inertia of your leg is greater when your leg is

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It would be bent, and your welcome

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3 years ago

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The thermal efficiency of a power cycle operating in a reversible manner is found to be 50%. Assuming that the same 2 thermal re

inna [77]

Answer:

Explanation:

The thermal efficiency of a Power cycle $\eta = \dfrac{Q_H -Q_c}{Q_H}$

where;

$\eta = 50\% = 0.5$

$Q_H = Heat \ flow \ from \ higher \ temperature$

$Q_c = Heat \ flow \ from \ lower \ temperature$

$0.5 = \dfrac{Q_H -Q_c}{Q_H}$

$0.5 Q_H = Q_H - Q_c$ --- (1)

$Q_c = 0.5 Q_H$ ---- (2)

The coefficient of performance is:

$COP_R = \dfrac{Q_c}{Q_H -Q_c}$

let replace the value of $Q_c = 0.5 Q_H$ in the above equation then;

$COP_R = \dfrac{0.5Q_H}{Q_H -0.5 Q_H}$

$COP_R = \dfrac{0.5Q_H}{0.5 Q_H}$

$COP_R = 1$

The

On the other hand, the heat pump

$COP_{HP} = \dfrac{Q_H}{Q_H -Q_c}$

By replacing equation (1) into the above equation; we have:

$COP_{HP} = \dfrac{Q_H}{0.5Q_{H}}$

$COP_{HP} = \dfrac{1}{0.5}$

$COP_{HP} =2$

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3 years ago

how much heat is lost when 71.07g of steam at 100 degrees condenses to form liquid water at 100 degrees

alukav5142 [94]

Answer:

Q = 160.36[kJ], is the heat lost.

Explanation:

This is a thermodynamic problem where we can find the latent heat of vaporization, at a constant temperature of 100 [°C].

We know that for steam at 100[C], the enthalpy is

$h_{gas} = 2675.6[kJ/kg]$

For liquid water at 100[C], the enthalpy is

$h_{water} = 419.17[kJ/kg]$

Therefore

$h_{g-w} = 2675.6-419.17 = 2256.43[\frac{kJ}{kg} ]$

The amout of heat is given by:

$Q=h_{g-w}*m\\ where:\\m = mass = 0.07107[kg] = 71.01[g]\\Q = heat [kJ]\\Q =2256.43*0.07107\\Q=160.36[kJ]$

3 0

3 years ago

The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of

Oksi-84 [34.3K]

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>

We have the expression for the initial velocity as,

$v=\sqrt{2gh}$

Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

$g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132$

Now, the velocity will become,

$v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s$

<h3>How to find the speed of the satellite?</h3>

As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

$v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s$

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

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3 0

2 years ago

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