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Roman55 [17]
3 years ago
15

How long does light take to travel from the sun to earth? heres the exact question: light travels at 300,000 km.s. The sun is a

total of 148 million kilometres from the earth. How long does it take light to travel from the sun to the earth
Physics
1 answer:
Artist 52 [7]3 years ago
4 0
I've heard it takes around 8 min
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What is drag? In what situations might it act on the force of gravity?
stepan [7]
Drag is passive, therefore it does not act on gravity. Drag is a mechanical resistance to motion. If the motion in question was induced by gravity drag it can impede that motion, but has no effect on the gravity. 

hope this helps!
7 0
3 years ago
Read 2 more answers
Helppp please i think I know the answer but i want to be sure
Vadim26 [7]

Answer:  D) 0-3 seconds

Explanation:

5 0
1 year 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
Two factors that regulate (control) glandular secretion.
antiseptic1488 [7]

Answer:

The factors include age and puberty

Explanation:

Glandular secretion release chemicals such as hormones in response to the body’s metabolic needs.

As an individual ages , the metabolic rate of the body also reduces . This is due to the stress and ageing of the cells of the body. This explains why glandular secretion is optimal with young people and Lower in older people. It also explains why the immune system of a young person is mostly stronger than older people.

Puberty is another factor which affects glandular secretion as during puberty there is usually a high amount of hormonal changes due to high levels of secretions of some hormones. These hormones could however inhibit the other glandular secretions.

6 0
3 years ago
The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasin
Ganezh [65]

Answer:

The current in the circuit decrease slowly .

Explanation:

Given as :

For the electrical circuit

The voltage V in the circuit is slowly decreasing

The resistance R of the resistor slowly increasing after heating

Now, From Ohm's Law

Voltage is directly proportional to the flow of current through circuit

I.e V ∝ I

Or. V = R × I

where R is the proportionate constant and this is the resistance of the resistor

whose property is to opposes the flow of current in the circuit

So, If R value more then current I reduces in the circuit

∵ Here in the circuit ,  The resistance is slowly increasing, so, current I is slowly decreasing .

Hence The current in the circuit decrease slowly . answer

3 0
3 years ago
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