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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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3 years ago
Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th
timama [110]

Answer:

Explanation:

Given

length of window h=2.9\ m

time Frame for which rock can be seen is \Delta t=0.134\ s

Suppose h is height above which rock is dropped

Time taken to cover h+2.9 is t_1

so using equation of motion

y=ut+\frac{1}{2}at^2

where  y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h  is

h=0+0.5\times g\times (t_2)^2---2

Subtract 1 and 2 we get

2.9=0.5g(t_1^2-t_2^2)

5.8=g(t_1+t_2)(t_1-t_2))

and from equation t_1-t_2=0.134\ s

so t_1+t_2=\frac{5.8}{9.8\times 0.134}

t_1+t_2=4.416\ s

and t_1=t_2+\Delta t

so t_2+\Delta t+t_2=4.416

2t_2+0.134=4.416

t_2=0.5\times 4.282

t_2=2.141\ s

substitute the value of t_2 in equation 2

h=0.5\times 9.8\times (2.141)^2

h=22.46\ m

                                                     

8 0
3 years ago
Which of the following is numerically the same as the specific gravity? Mass Weight Density Volume
max2010maxim [7]
The answer is density

4 0
3 years ago
Read 2 more answers
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
3 years ago
A butcher grinds 5 and 3/4 lb of meat then sells it for 2 and 2/3 pounds to the customer what is the maximum amount me that the
KiRa [710]

Answer:

The maximum amount of meat that the butcher can sell is  3\frac{1}{12}\:lb

Explanation:

The maximum amount can be found by taking the difference of mixed numbers.

5\frac{3}{4}-2\frac{2}{3}\\\\\mathrm{Subtract\:the\:numbers:}\:5-2=3\\\\\mathrm{Combine\:fractions:\:}\frac{3}{4}-\frac{2}{3}=\frac{1}{12}\\\\=3\frac{1}{12}\\

Best Regards!

8 0
3 years ago
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