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miskamm [114]
3 years ago
14

A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte

r one time constant? You are supposed to use a constant e can you please include what the value should be for this and all other constants.
Physics
1 answer:
ivanzaharov [21]3 years ago
5 0

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

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The frictional force is given by F = μmg 

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<span>Work done by frictional force = Fd = μmgd </span>

<span>Kinetic energy "lost" = 1/2 mv² </span>


<span>Fd = μmgd = 1/2 mv² </span>

<span>The m's cancel μgd = v² / 2 </span>



<span>d = v² / 2μg </span>

<span>d = 8² / 2(0.41)(9.8) </span>

<span>d = 32 / (0.41)(9.8) </span>

<span>d = 7.96 </span>

<span>Player slides 8 m . </span>



<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>



<span>d = v² / 2μg </span>




<span>= 4² / 2(0.46)(9.8) </span>

<span>= 8 / (0.46)(9.8) </span>

<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
3 0
3 years ago
Read 2 more answers
A mass of 10 g of oxygen fill a weighted piston–cylinder device at 20 kPa and 110°C. The device is now cooled until the tempe
mezya [45]

Answer:

The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

Explanation:

Given that,

Mass of oxygen = 10 g

Pressure = 20 kPa

Initial temperature = 110°C

Final temperature = 0°C

We need to calculate the change of the volume of the device during this cooling

Using formula of change volume

\Delta V=V_{2}-V_{1}

\Delta V=\dfrac{mR}{P}(T_{2}-T_{1})

Put the value into the formula

\Delta V=\dfrac{0.3125\times0.0821}{2.0265\times10^{9}}(383-273)

\Delta V=14.297\ L

\Delta V=14.3\times10^{-3}\ m^3

Hence, The change of the volume of the device during this cooling is 14.3\times10^{-3}\ m^3

6 0
3 years ago
An 820 N Marine in basic training climbs a 12.0-m vertical rope at a constant speed in 8.00 s. What is her power output
sweet-ann [11.9K]

Answer:

1230 W

Explanation:

P = \frac{W}{t} = \frac{Force * distance}{time} = \frac{820 N * 12.0 m}{8.00 s} = 1230 Watts

6 0
3 years ago
A 38.5kg man is in an elevator accelerating downward. A normal force of 343n pushes up on him. what is his acceleration?
alexira [117]

Answer:

<h3>The answer is 8.91 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

f is the force

m is the mass

From the question we have

a =  \frac{343}{38.5}  =  \frac{98}{11}  \\  = 8.909090...

We have the final answer as

<h3>8.91 m/s²</h3>

Hope this helps you

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2 years ago
A skydiver jumps from an airplane and falls freely for 5 seconds. If the sky diver's velocity increases by 49 m/s during that ti
Sergeeva-Olga [200]

Acceleration  =  (change in speed)  /  (time for the change)

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                       =       (49 / 5)  m/s  /  s

                       =            9.8 m/s² 
8 0
2 years ago
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