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oksian1 [2.3K]
3 years ago
11

You create a plot of voltage (in V) vs. time (in s) for an RC circuit as the capacitor is charging, where V=V_{0} \cdot \left(1-

e^{ \frac{-\left(t\right)}{RC} } \right). You curve fit the data using the inverse exponent function Y=A \cdot \left(1- e^{-\left(Cx\right)} \right)+B and LoggerPro gives the following values for A, B, and C. A = 4.211 ± 0.4211 B = 0.1699 ± 0.007211 C = 1.901 ± 0.2051 What is the time constant for and its uncertainty?
Physics
1 answer:
creativ13 [48]3 years ago
3 0

Answer:

The time constant and its uncertainty is t ± Δt = 0.526 ± 0.057 s

Explanation:

If we make a comparison we have to:

y = A*(1-e^-(C*x)) + B

If the time remains constant we have to:

t = R*C = 1/C

In this way we calculate the time constant and its uncertainty. this will be equal to:

t ± Δt = (1/1.901) ± (0.2051/1.901)*(1/1.901) = 0.526 ± 0.057 s

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because they are found freely in nature uncombined so they are highly reactive with other elements

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Any object that is given any initial velocity and which follows a path due to gravitational force acting on it and by the fricti
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Any object that is given any initial velocity and which follows a path due to gravitational force acting on it and by the frictional resistance of the atmosphere is called a projectile. This is because the object is projected and not influenced by anything except gravity.

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jeremie ran around the track at the ymca for 2 hours. when he was done he figured that he had traveled 20 kilometers. what was h
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(1) Speed is the ratio of the total distance covered by the object and the total time it takes for him to finish it. 

    Speed = distance / time 

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   Speed = 20 kilometers / 2 hours 
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3 years ago
Using the American Engineering system of units (AES), a) Calculate the weight of a 170.5 lbm person on the surface of the earth,
Soloha48 [4]

Answer:

a) the weight of the person is 170.5 lbf

b) weight of the astronaut on the moon is 28.2 lbf

Explanation:

Given the data in the question;

a)

we know that;

weight on the surface of the earth = mg_{earth

given that m = 170.5 lbm and g = 32.174 ft/s²

we substitute

weight on the surface of the earth =  170.5 lbm × 32.174 ft/s²

= 5485.667 lbm-ft/s²

1 lbf = 32.174 lbm-ft/s²

so

weight on the surface of the earth = (5485.667 / 32.174) lbf

weight on the surface of the earth = 170.5 lbf

Therefore, the weight of the person is 170.5 lbf

b)

given that;

weight on the surface of the earth = mg_{moon

m = 170.5 lbm and g = 5.32 ft/s²

weight on the surface of the earth = 170.5 lbm × 5.32 ft/s²

= 907.06 lbm-ft/s²

1 lbf = 32.174 lbm.ft/s²

weight on the surface of the earth = ( 907.06 / 32.174 ) lbf

weight on the surface of the earth = 28.2 lbf

Therefore, weight of the astronaut on the moon is 28.2 lbf

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3 years ago
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