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Andrew [12]
3 years ago
5

Sarah's group designed this oven and eventually melted the

Physics
1 answer:
torisob [31]3 years ago
5 0

Answer:

b

Explanation:

because i don't he asking me

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Can you help me answer this?
Pavel [41]
The answer is D. If you aren't consistent with your drop positions, then your data may be invalid. To be frank: it basically screws over the experiment.
5 0
3 years ago
How do you increase the potential energy of an apple
m_a_m_a [10]
To increase the potential energy of any object, not limited to an apple, you have to hold it higher.
5 0
3 years ago
The op amp in this circuit is ideal. R3 has a maximum value of 100 kΩ and σ is restricted to the range of 0.2 ≤ σ ≤ 1.0. a. Calc
Firlakuza [10]

I have attached the circuit image missing in the question.

Answer:

A) The range of vo is; -6.6V≤ vo ≤-1V

B) σ = 0.1861

Explanation:

A) First of all, Let VΔ be the voltage from the potentiometer contact to the ground.

Thus; [(0 - vg)/(2000)] +[(0 - vΔ)/(50,000)] = 0

So, [(- vg)/(2000)] +[(- vΔ)/(50,000)] = 0

Simplifying further; -25 vg - vΔ = 0

From the question, vg = 40mV = 0.04 V

So - 25(0.04) = vΔ

So: vΔ = - 1 V

Now, [vΔ/(σRΔ)] + [(vΔ - 0)/(50,000)] + [(vΔ - vo)/((1 - σ)RΔ))] = 0

So, multiplying each term by RΔ to get; [vΔ/(σ)] + [(vΔ x RΔ)/(50,000)] + [(vΔ - vo)/((1 - σ))] = 0

So RΔ = 100kΩ or 100,000Ω from the question.

So, substituting for RΔ, we get,

[vΔ/(σ)] + [2vΔ] + [(vΔ - vo)/((1 - σ))] = 0

Let's put the value of - 1 for vΔ as gotten before.

So, ( - 1/σ) - 2 + [(-1 - vo)/(1 - σ)] = 0

Now let's make vo the subject of the equation to get;

-1 - vo = (1 - σ)[2 + (1/σ)]

-1 - vo = 2 - 2σ + (1/σ) - 1

-vo = 1 + 2 - 2σ + (1/σ) - 1

-vo = 2 - 2σ + (1/σ)

vo = - 1 (2 - 2σ + (1/σ))

When σ = 0.2; vo = - 1(2 - 0.4 + 5) =

- 1 x 6.6 = - 6.6V

Also when σ = 1;

vo = - 1(2 - 2 + 1) = - 1V

Therefore, the range of vo is;

- 6.6V ≤ vo ≤ - 1V

B) it will saturate at vo = - 7V

So, from;

vo = - 1 (2 - 2σ + (1/σ))

-7 = - 1 (2 - 2σ + (1/σ))

Divide both sides by (-1)

7 = (2 - 2σ + (1/σ))

Now, subtract 2 from both sides to get; 5 = - 2σ + (1/σ)

Multiply each term by α to get;

5σ = - 2σ^(2) + 1

So 2σ^(2) + 5σ - 1 = 0

Solving simultaneously and picking the positive value , we get σ to be approximately 0.1861

8 0
3 years ago
Does a car have antennae? what senses does it have
Ilia_Sergeevich [38]

Answer:

Explained below.

Explanation:

Yes cars have antennas.

Now there could be an antenna for listening to radio stations which could be regular radio stations or even satellite radio stations.

Also, there could be antennas used for GPS navigation or some form of communication with other vehicles that possess that type of antenna.

4 0
3 years ago
The intensity level is 65 db at a distance 5.00 m from a barking dog. What would be a reasonable estimate for the intensity leve
BARSIC [14]

Answer:

68 db

Explanation:

Since now instead of one two dogs are barking simultaneously close to each other, therefore we take n =2.

Ignoring interference effects, the barking of two dogs result in a higher level of intensity which is given by,

β(db)=10×㏒(2)

=3 db

So, a reasonable estimate for the raised Intensity Level is: 65db+3db = 68db

7 0
3 years ago
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