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4 years ago

A battery can be modeled as a source of emf in series, with both a resistance of 10Ωand an inductive reactance of 5

Physics

1 answer:

Fiesta28 [93]4 years ago

3 0

Answer:

Explanation:

For circuit in resonance , inductive and capacitive reactance should be equal. Since in the given circuit , these two are not equal, the circuit is not in resonance. It does not depend upon voltage provided.

option b is correct.

With this load, the fraction of the average power, put out by the source of emf, delivered to the load can be calculated as follows

Power delivered to resistor will be 2/3 rd of total power delivered by source because resistance has value twice that of reactance of capacitor. So the correct option is .7

option D ) is correct.

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The current definition of the standard kilogram of mass is based onA) the mass of the earth.B) the mass of the sun.C) the mass a

Neko [114]

The kilogram SI unit is based off an original prototype that is kept in France. The prototype itself is based off the mass of water under special conditions for temperature as well as pressure.

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3 years ago

A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de

oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

$\eta = \frac{W}{Q_{in}}$

where

W is the useful work done by the engine

$Q_{in}$ is the heat in input to the machine

In this problem, we have:

$Q_{in}=600 kJ$ is the heat absorbed

$W=600 kJ-175 kJ=425 kJ$ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)

So, the efficiency is

$\eta = \frac{425 kJ}{600 kJ}=0.71 = 71\%$

(a) $737^{\circ}C$

The efficiency of an engine can also be rewritten as

$\eta = 1-\frac{T_C}{T_H}$

where

$T_C$ is the absolute temperature of the cold sink

$T_H$ is the temperature of the source

In this problem, the temperature of the sink is

$T_C = 20^{\circ}C + 273=293 K$

So we can re-arrange the equation to find the temperature of the source:

$T_H = \frac{T_C}{1-\eta}=\frac{293 K}{1-0.71}=1010 K\\T_H = 1010 K - 273=737^{\circ}C$

7 0

3 years ago

A projectile was launched horizontally with a velocity of 388 m/s, 2.89 m above the ground. How long did it take the projectile

tamaranim1 [39]

Answer:

Explanation:

Given

Velocity = 388m/s

Height S = 2.89m

Required

Time

Using the equation of motion

S =ut+1/2gt²

2.89 = 388t+1/2(9.8)t²

2.89 = 388t+4.9t²

Rearrange

4.9t²+388t-2.89 =0

Factorize

t = -388±√388²-4(4.9)(2.89)/2(4.9)

t= -388±√(388²-56.644)/9.8

t = -388±387.93/9.8

t =0.073/9.8

t = 0.00744 seconds

6 0

3 years ago

In a certain ideal heat engine, 10.00 kJ of heat is withdrawn from the hot source at 273 K and 3.00 kJ of work is generated. Wha

k0ka [10]

Answer:

Temperature of the sink will be 191.1 K

Explanation:

We have given that heat withdrawn form the source = 10 KJ

Work done = 3 KJ

We know that efficiency is given by

$\eta =\frac{work\ done}{heat\ withdrawn}=\frac{3}{10}=0.3$

Higher temperature is given by $T_1=273K$

We have to find the lower temperature $T_2$

We know that efficiency is also given by

$1-\frac{T_2}{T_1}=\frac{T_1-T_2}{T_1}$

So $\frac{273-T_2}{273}=0.3$

$T_2=191.1K$

So temperature of the sink will be 191.1 K

5 0

3 years ago

Read 2 more answers

Six members of a synchronized swim team wear earplugs to protect themselves against water pressure at depths, but they can still

tangare [24]

Answer:

The sound travels differently in different medium according the density of the medium.

Explanation:

The sound travels faster in dense medium and can be heard by the vibration of the bone present in the ear. The ear plugs reduce the sound intensity in both medium water and on land (air).

In air the sound is not heard properly due to the earplugs that stops the as the vibration are not able to produce as sound is not able to reach to middle ear, but Navy researchers have discovered that sound under water is heard by the bone present behind the ear, vibrations mastoid.

3 0

4 years ago