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alekssr [168]
3 years ago
13

In which case would electrical potential energy be built up and stored in the electric field?

Physics
1 answer:
olasank [31]3 years ago
8 0

Answer: b) A positive charge is moved toward a positive charge.

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0<br> What is 33 C in absolute temperature? hs
AysviL [449]

Answer:

33 Celsius is 306.15 in absolute temperature

7 0
2 years ago
Is a Joule the same as a Kelvin?
earnstyle [38]
No,because a Kevin is 7.242971666663E+22 times Smaller than a Joule.
7 0
3 years ago
Read 2 more answers
A 5,000 kg truck moving at 8 m/s has the same momentum as a 2,500 kg car. What is the velocity of the car?
rodikova [14]

Answer:

16 m/s.

Explanation:

The following data were obtained from the question:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Mass of car = 2500 kg

Velocity of car =..?

Next, we shall determine the momentum of the truck. This can be obtained as follow:

Mass of truck = 5000 Kg

Velocity of truck = 8 m/s

Momentum of truck =.?

Momentum = mass × velocity

Momentum = 5000 × 8

Momentum of the truck = 40000 Kg.m/s

Finally, we shall determine the velocity of the car as follow:

From the question given above, we were told that the car and truck has the same momentum.

This implies that:

Momentum of the truck = momentum of car = 40000 Kg.m/s

Thus, the velocity of the car can be obtained as shown below:

Mass of car = 2500 kg

Momentum of the car = 40000 Kg.m/s

Velocity of car =..?

Momentum = mass × velocity

40000 = 2500 × velocity

Divide both side by 2500

Velocity = 40000/2500

Velocity = 16 m/s

Therefore, the velocity of the car is 16 m/s.

5 0
3 years ago
Where is the value of kinetic energy equal to zero
bearhunter [10]

Answer:

if the object is not in motion

Explanation:

7 0
3 years ago
Read 2 more answers
to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
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