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PolarNik [594]
4 years ago
14

A heating element having a resistance (at its operating temperature) of 148 Ω is connected to a battery having an emf of 523 V a

nd unknown internal resistance. It is found that heat energy is being generated in the resistance of the heating element at a rate of 66.0 W. What is the rate at which heat energy is being generated in the internal resistance of the battery?
Physics
2 answers:
Fudgin [204]4 years ago
7 0

One formula for the power dissipated by a resistor is

Power = (current)² · (resistance)

I have a feeling that this would be a handy formula to use to solve this problem.

66 W = (current)² · (148 Ω)

Current = √(66/148)

Current = 0.668 Ampere

The total power being delivered by the battery is . . .

Power = (voltage) · (current)

Power = (523 volts) · (0.668 Amp)

Total battery power = 349.4 Watts

66 Watts is being dissipated by the heating element.  The rest of the humongous power delivered by the battery is being dissipated by the battery's own internal resistance.

Power = (349.4 - 66)

<em>Power = 283.4 watts</em>

<em></em>

As a working electrical engineer, I'm required to pound on the desk and protest this expensive and unhealthy situation.  

-- The battery is enormous ... 523 V ! ... and must weigh a ton.  

-- The battery itself is using 81% of the power it's being used to deliver.  

-- And all this is just to operate a measly 66W heater !  

-- You can probably get just as much heat out of the heater in the truck you'll need to shlep this battery around to where the heat is needed.

vladimir2022 [97]4 years ago
4 0

Answer:

The correct answer is 283 watts

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You observe two stars over the course of a year (or more) and find that both stars have measurable parallax. (1 arc second is 1/
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Answer:

Star X is much closer since it is at a distance 1 parsec from the Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth. This angle is gotten when the position of the object is measured in January and then in July according to the configuration of the Earth with respect to the Sun in those months.

The distance between the Earth and the Sun is 150000000 Km. That distance is also known as an astronomical unit (1AU).

The parallax angle can be defined in the following way:

\tan{p} = \frac{1AU}{d}    

Where d is the distance to the star.

p('') = \frac{1}{d}    (1)  

Equation (1) can be rewritten in terms of d:

d(pc) = \frac{1}{p('')} (2)

Equation (2) represents the distance in a unit known as parsec (pc).

Case of Star X (p('') = 1):

Using equation 2 the distance of star X can be known:

d(pc) = \frac{1}{1}

d(pc) = 1 pc

So, star X is at 1 parsec from Earth.

Case of Star Y (p('') = \frac{1}{2}):

d(pc) = \frac{1}{(\frac{1}{2})}

d(pc) = 2 pc

So, star Y is at 2 parsecs from the Earth.

Hence, star X is much closer.

Reminder:

Notice that in equation 2 the distance is inversely proportional to the parallax angle, so if the parallax angle decreases, the distance increases.

5 0
3 years ago
Suppose a piece of food is on the edge of a rotating microwave oven plate. Does it experience nonzero tangential acceleration, c
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Answer:

Explanation:

(a) When the plate starts to spin:

Its angular velocity increases, so the angular acceleration is non zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(b) When the plate rotates at constant angular velocity:

Its angular velocity is constant so the angular acceleration is zero. As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

(c) When the plate sows to halt:

Its angular velocity decreases, so the angular acceleration is non zero( but negative). As the direction of velocity keeps on changing every instant so the linear acceleration is also non zero.

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A 20-kilogram mass is traveling with a velocity of 3 m/s. What is the object's kinetic energy?
jolli1 [7]

solution

Given

mass= 20kg

velocity =3m/s

K.E=?

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½×20×9

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Answer:

\frac{729\frac{m^{2} }{s^{2} } }{83m}≈8.8\frac{m}{s^{2} }

Explanation:

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