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3 years ago

Use V=Voe^-t/RC and T=RC to derive T1/2=Tln2

Physics

1 answer:

Ksju [112]3 years ago

3 0

Enclosed is some guidance algebra.I find this q a little confusing. It quotes "RC" which usually makes me think of electrical circuits and time constants based on converting calculating RC value and equating that to t for one time constant then 2RC for two time constants etc. The theory being that after 5 time constants - 5RC - a circuit is stable. BUT, this q then goes on to mention HALF LIFE. The curves for both half life and time constant are both exponential, as in the number e to the power of something, but the algebra is slightly different. I hope my algebra is ok.

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Discuss the merits and demerits of various theories of the formation of galaxies.

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3 years ago

Does specific heat of a substance depend on its temperature?

lara [203]

Answer:

temperature

Explanation:

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3 years ago

What is a therapuetic modality in psychology?

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2 years ago

Read 2 more answers

Two tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.8×106N , one an angle 11degrees west of north an

sp2606 [1]

Answer: 2,9 ×10¹⁰J.

Explanation:

1. The work is a measure of energy and is calculated as the product of the displacement times the parallel force to such displacement.

That means that the only components of the force that contribute to work are those that result parallel to the displacement.

2. Since it is given that the two tugboats "pull the tanker a distance 0.83km toward the north", that is the displacement, and you have to calculate the net force toward the north.

3. Tugboat #1.

a) Force magnitude: F₁ = 1.8×10⁶N

b) Angle: α = 11° West of North

c) North component of the force F₁: Fy₁ = F₁cos(α) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N

4. Tugboat #2:

a) Force magnitude: F₂ = 1.8×10⁶N

b) Angle: = 11° East of North

c) North component of the force F₂: Fy₂ = F₂cos(β) = 1.8×10⁶N × cos(11°) = 1.77×10⁶N =

5. Total net force, Fn:

Fn = Fy₁ + Fy₂ = 1.77×10⁶N + 1.77×10⁶N = 3.54×10⁶N

6. Work, W:

Displacement, d = 0.83 km = 8,300 m

W = Fn×d = 3.54×10⁶N×8,300m = 29,000 ×10⁶J = 2,9 ×10¹⁰J

The answer is rounded to two significant figures because both data, Force and displacement, have two significant figures.

7 0

3 years ago

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0

3 years ago